OP Malhotra Class 12 Maths Solutions Chapter 28 Linear Programming Exercise 28 (C)

S Chand Class 12 ICSE Maths Solutions Chapter 28 Linear Programming Ex 28(c)

 

Question 1. A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of the first machine is 12 hours and that of second machine is 9 hours. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on the second machine. Each unit of product A is sold at a profit of Rs 5 and B at a profit of Rs 6, find the maximum profit graphically.
Answer: Solution:
Given data can be arranged in tabular form as under :

Let \(x\) units of product A and \(y\) units of product B are produced by the manufacturer.
Since each unit of product A is sold at a profit of Rs 5 and product B at a profit of Rs 6, the total profit earned by the manufacturer by selling \(x\) units of product A and \(y\) units of product B will be \(5x + 6y\).
Let \(Z\) be the total profit. Therefore, \(Z = 5x + 6y\), and our goal is to maximise \(Z\).
The corresponding constraints are:
Machine I constraint: \(3x + 2y \leq 12\)
Machine II constraint: \(3x + y \leq 9\)
Number of units cannot be negative: \(x \geq 0\); \(y \geq 0\)
To find the region for \(3x + 2y \leq 12\), we first draw the line \(3x + 2y = 12\). This line intersects the coordinate axes at A (4, 0) and B (0, 6). Since (0, 0) satisfies the inequality (\(3(0) + 2(0) = 0 \leq 12\)), the region containing (0, 0) is part of the solution set.
For the region \(3x + y \leq 9\), we draw the line \(3x + y = 9\). This line intersects the coordinate axes at C (3, 0) and D (0, 9). Since (0, 0) satisfies the inequality (\(3(0) + 0 = 0 \leq 9\)), the region containing (0, 0) is part of the solution set.
The conditions \(x \geq 0\) and \(y \geq 0\) represent the first quadrant of the XOY plane.
The two lines \(3x + 2y = 12\) and \(3x + y = 9\) intersect at point P (2, 3). To find this point, we can solve the system of equations:
1. \(3x + 2y = 12\)
2. \(3x + y = 9\)
Subtract equation (2) from equation (1):
\( (3x + 2y) - (3x + y) = 12 - 9 \)
\( y = 3 \)
Substitute \(y = 3\) into equation (2):
\( 3x + 3 = 9 \)
\( 3x = 6 \)
\( x = 2 \)
So, the intersection point is (2, 3).
The shaded region represents the feasible region, and its corner points are O (0, 0), C (3, 0), P (2, 3), and B (0, 6). We now evaluate \(Z\) at these corner points:

The maximum value of \(Z\) is 36, which occurs at point B (0, 6). This means \(x = 0\) and \(y = 6\).
Therefore, the maximum profit of Rs 36 is earned when 0 units of product A and 6 units of product B are produced. This approach involves checking the profit at each vertex of the feasible region, which is a standard method in linear programming.

In simple words: First, we list all the production information like machine hours and profit. Then, we use this to make mathematical rules (constraints) and a profit equation. We draw these rules on a graph to find the area where all rules are followed. Finally, we check the profit at the corners of this area to find the highest possible profit.

🎯 Exam Tip: Always clearly define your variables (x and y) and the objective function (Z). Drawing the feasible region correctly and identifying all corner points is crucial for full marks.

 

Question 2. A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 calories. Two foods X and Y are available at a cost of Rs 4 and Rs 3 per unit respectively. One unit of food X contains 200 units of vitamins, 1 unit of minerals and 40 calories. One unit of food Y contains 100 units of vitamins, 2 units of minerals and 40 calories. Find what combinations of foods X and Y should be used to have least cost, satisfying the requirements.
Answer: Solution:
Let \(x\) units of food X and \(y\) units of food Y are used to minimize the cost.
The costs of one unit of food X and Y are Rs 4 and Rs 3 per unit respectively. So, the cost for \(x\) units of food X and \(y\) units of food Y will be \(4x\) and \(3y\) respectively.
Let \(Z\) be the total cost. Then, the objective function is \(Z = 4x + 3y\), and our aim is to minimize \(Z\).
Now, let's formulate the constraints based on the minimum requirements:
1. Vitamin Requirement:
One unit of food X contains 200 units of vitamins, and one unit of food Y contains 100 units. The minimum requirement is 4000 units.
So, \(200x + 100y \geq 4000\)
Dividing by 100, we get: \(2x + y \geq 40\) (First constraint)
2. Mineral Requirement:
One unit of food X contains 1 unit of minerals, and one unit of food Y contains 2 units. The minimum requirement is 50 units.
So, \(x + 2y \geq 50\) (Second constraint)
3. Calorie Requirement:
One unit of food X contains 40 calories, and one unit of food Y contains 40 calories. The minimum requirement is 1400 calories.
So, \(40x + 40y \geq 1400\)
Dividing by 40, we get: \(x + y \geq 35\) (Third constraint)
4. Non-negativity Constraint:
The quantity of food cannot be negative, so \(x \geq 0\) and \(y \geq 0\).
So, the complete mathematical formulation of the Linear Programming Problem (LPP) is:
Minimize \(Z = 4x + 3y\)
Subject to constraints:
\(2x + y \geq 40\)
\(x + 2y \geq 50\)
\(x + y \geq 35\)
\(x \geq 0, y \geq 0\)
To determine the feasible region, we examine each inequality:
* For \(2x + y \geq 40\): The line \(2x + y = 40\) meets the coordinate axes at A\(_{1}\)(20, 0) and B\(_{1}\)(0, 40). Since (0, 0) does not satisfy the inequality (\(2(0) + 0 = 0 \not\geq 40\)), the solution set is the region not containing the origin.
* For \(x + 2y \geq 50\): The line \(x + 2y = 50\) meets the coordinate axes at C\(_{1}\)(50, 0) and B\(_{2}\)(0, 25). Since (0, 0) does not satisfy the inequality (\(0 + 2(0) = 0 \not\geq 50\)), the solution set is the region not containing the origin.
* For \(x + y \geq 35\): The line \(x + y = 35\) meets the coordinate axes at E\(_{1}\)(35, 0) and B\(_{3}\)(0, 35). Since (0, 0) does not satisfy the inequality (\(0 + 0 = 0 \not\geq 35\)), the solution set is the region not containing the origin.
* \(x \geq 0, y \geq 0\) represents the first quadrant of the xy-plane.
The lines \(x + 2y = 50\) and \(x + y = 35\) intersect at H\(_{1}\)(20, 15).
To find H\(_{1}\), solve:
1. \(x + 2y = 50\)
2. \(x + y = 35\)
Subtract equation (2) from equation (1):
\( (x + 2y) - (x + y) = 50 - 35 \)
\( y = 15 \)
Substitute \(y = 15\) into equation (2):
\( x + 15 = 35 \)
\( x = 20 \)
So, H\(_{1}\) is (20, 15).
The lines \(2x + y = 40\) and \(x + y = 35\) intersect at G\(_{1}\)(5, 30).
To find G\(_{1}\), solve:
1. \(2x + y = 40\)
2. \(x + y = 35\)
Subtract equation (2) from equation (1):
\( (2x + y) - (x + y) = 40 - 35 \)
\( x = 5 \)
Substitute \(x = 5\) into equation (2):
\( 5 + y = 35 \)
\( y = 30 \)
So, G\(_{1}\) is (5, 30).
The unbounded shaded region C\(_{1}\)H\(_{1}\)G\(_{1}\)B\(_{1}\) represents the feasible region with corner points C\(_{1}\)(50,0), H\(_{1}\)(20, 15), G\(_{1}\)(5, 30), B\(_{1}\)(0, 40).
Now, we evaluate \(Z\) at these corner points:

The smallest value of \(Z\) is 110 at \(x = 5, y = 30\).
Since the feasible region is unbounded, we need to check if this minimum value is truly the lowest. We do this by drawing the line \(4x + 3y = 110\) and checking if the open half-plane \(4x + 3y < 110\) has any common points with the feasible region.
The open half-plane \(4x + 3y < 110\) has no points in common with the feasible region, which confirms that 110 is indeed the minimum value. This means the combination that costs the least is 5 units of food X and 30 units of food Y. The minimum cost is Rs 110.

In simple words: We need to mix two foods to meet vitamin, mineral, and calorie needs with the lowest cost. We write down rules (constraints) for each nutrient and the total cost. Then, we draw these rules on a graph to find the smallest cost at the corners of the allowed area. This helps us find the cheapest way to get all the required nutrients.

🎯 Exam Tip: When dealing with "at least" requirements, the inequalities will use "≥". For unbounded regions, always check the objective function in the open half-plane to confirm the true minimum or maximum.

 

Question 3. A small firm manufactures gold rings and chains. The combined number of rings and chains manufactured per day is atmost 24. It takes one hour to make a ring and half an hour to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs 300 and on a chain Rs 190, how many of each should be manufactured daily so as to maximize the profit.
Answer: Solution:
Let \(x\) be the number of gold rings and \(y\) be the number of chains manufactured by the firm each day to get the maximum profit.
The profit on each ring is Rs 300, and on each chain is Rs 190. So, the profit on \(x\) units of rings and \(y\) units of chains will be Rs \(300x\) and Rs \(190y\) respectively.
Let \(Z\) be the total profit. Then \(Z = 300x + 190y\), and we need to maximize this.
Now, let's establish the constraints:
1. Combined Production Constraint:
The combined number of rings and chains manufactured per day is at most 24.
So, \(x + y \leq 24\) (First constraint)
2. Time Constraint:
It takes 1 hour to make a ring and 30 minutes (0.5 hours or 1/2 hour) to make a chain. The maximum time available is 16 hours.
Converting everything to minutes: 1 hour = 60 minutes, 0.5 hours = 30 minutes, 16 hours = 960 minutes.
So, \(60x + 30y \leq 960\)
Dividing by 30, we get: \(2x + y \leq 32\) (Second constraint)
3. Non-negativity Constraint:
The number of rings and chains cannot be less than zero.
So, \(x \geq 0, y \geq 0\).
Hence, the mathematical formulation of the LPP is:
Maximize \(Z = 300x + 190y\)
Subject to constraints:
\(x + y \leq 24\)
\(2x + y \leq 32\)
\(x \geq 0, y \geq 0\)
To determine the feasible region, we analyze each inequality:
* For \(2x + y \leq 32\): The line \(2x + y = 32\) meets the coordinate axes at C(16,0) and D(0,32). Since (0,0) satisfies the inequality (\(2(0) + 0 = 0 \leq 32\)), the region containing the origin is the solution set.
* For \(x + y \leq 24\): The line \(x + y = 24\) meets the coordinate axes at A(24,0) and B(0,24). Since (0,0) satisfies the inequality (\(0 + 0 = 0 \leq 24\)), the region containing the origin is the solution set.
* \(x \geq 0, y \geq 0\) represents the first quadrant of the xy-plane.
The point E(8, 16) is obtained by solving \(2x + y = 32\) and \(x + y = 24\) simultaneously.
To find E, solve:
1. \(2x + y = 32\)
2. \(x + y = 24\)
Subtract equation (2) from equation (1):
\( (2x + y) - (x + y) = 32 - 24 \)
\( x = 8 \)
Substitute \(x = 8\) into equation (2):
\( 8 + y = 24 \)
\( y = 16 \)
So, E is (8, 16).
The shaded region OCEB represents the feasible region, and its corner points are O (0,0), C (16,0), E (8,16), and B (0,24).
Now, we evaluate \(Z\) at these corner points:

The maximum value of \(Z\) is 5440, which occurs at \(x = 8\) and \(y = 16\).
Therefore, the firm should manufacture 8 rings and 16 chains to achieve the maximum profit of Rs 5440. This ensures that all daily limits are met while making the most money. The calculations show how each combination of products contributes to the total profit.

In simple words: To make the most profit, the firm should make 8 gold rings and 16 chains. This specific combination fits within the total number of items and the machine time available each day, giving the highest profit of Rs 5440.

🎯 Exam Tip: Always represent time constraints in the same unit (e.g., all hours or all minutes) to avoid calculation errors. Remember that the "at most" condition translates to a "≤" inequality.

 

Question 4. If a young man rides his motor cycle at 25 km/hr, he has to spend Rs 2 per km on petrol, if he rides at a faster speed of 40 km/hr, the petrol cost increases to Rs 5 per km. He has Rs 100 to spend on petrol and wishes to find maximum distance he can travel within one hour. Express this as a linear programming problem and then solve it.
Answer: Solution:
Let the young man drive his scooter \(x\) km at a speed of 25 km/hr and \(y\) km at a speed of 40 km/hr.
The total distance traveled will be \(Z = x + y\), and the aim is to maximize \(Z\).
Now, let's set up the constraints:
1. Petrol Cost Constraint:
The cost is Rs 2 per km at 25 km/hr and Rs 5 per km at 40 km/hr. He has Rs 100 to spend.
So, \(2x + 5y \leq 100\) (First constraint)
2. Time Constraint:
Time taken to travel \(x\) km at 25 km/hr is \( \frac{x}{25} \) hours.
Time taken to travel \(y\) km at 40 km/hr is \( \frac{y}{40} \) hours.
He has 1 hour to travel in total.
So, \( \frac{x}{25} + \frac{y}{40} \leq 1 \)
To remove fractions, multiply by the LCM of 25 and 40, which is 200:
\( 200 \times \frac{x}{25} + 200 \times \frac{y}{40} \leq 200 \times 1 \)
\( 8x + 5y \leq 200 \) (Second constraint)
3. Non-negativity Constraint:
Distance cannot be less than zero.
So, \(x \geq 0, y \geq 0\).
Hence, the mathematical formulation of the LPP is:
Maximize \(Z = x + y\)
Subject to constraints:
\(2x + 5y \leq 100\)
\(8x + 5y \leq 200\)
\(x \geq 0, y \geq 0\)
To determine the feasible region, we analyze each inequality:
* For \(2x + 5y \leq 100\): The line \(2x + 5y = 100\) meets the coordinate axes at A\(_{1}\)(50, 0) and B\(_{1}\)(0, 20). Since (0, 0) satisfies the inequality (\(2(0) + 5(0) = 0 \leq 100\)), the region containing the origin is the solution set.
* For \(8x + 5y \leq 200\): The line \(8x + 5y = 200\) meets the coordinate axes at A\(_{2}\)(25, 0) and B\(_{2}\)(0, 40). Since (0, 0) satisfies the inequality (\(8(0) + 5(0) = 0 \leq 200\)), the region containing the origin is the solution set.
* \(x \geq 0, y \geq 0\) represents the first quadrant of the region.
The point P\( \left(\frac{50}{3}, \frac{40}{3}\right) \) is obtained by solving \(8x + 5y = 200\) and \(2x + 5y = 100\) simultaneously.
To find P, solve:
1. \(8x + 5y = 200\)
2. \(2x + 5y = 100\)
Subtract equation (2) from equation (1):
\( (8x + 5y) - (2x + 5y) = 200 - 100 \)
\( 6x = 100 \)
\( x = \frac{100}{6} = \frac{50}{3} \)
Substitute \(x = \frac{50}{3}\) into equation (2):
\( 2\left(\frac{50}{3}\right) + 5y = 100 \)
\( \frac{100}{3} + 5y = 100 \)
\( 5y = 100 - \frac{100}{3} = \frac{300 - 100}{3} = \frac{200}{3} \)
\( y = \frac{200}{15} = \frac{40}{3} \)
So, P is \( \left(\frac{50}{3}, \frac{40}{3}\right) \).
The corner points are O (0,0), A\(_{2}\)(25,0), P\( \left(\frac{50}{3}, \frac{40}{3}\right) \), and B\(_{1}\)(0, 20).
Now, we evaluate \(Z\) at these corner points:

The maximum value of \(Z\) is 30, which occurs at \(x = \frac{50}{3}\) and \(y = \frac{40}{3}\).
The distance covered at 25 km/hr is \( \frac{50}{3} \) km, and at 40 km/hr is \( \frac{40}{3} \) km. The maximum total distance traveled is 30 km. This means by combining driving at different speeds, the young man can cover the greatest distance while staying within his budget and time limit.
In simple words: The young man wants to travel the furthest while spending no more than Rs 100 on petrol and finishing in one hour. We set up equations for his cost and time based on his two speeds. By graphing these rules, we find the best combination of distances at each speed to cover a maximum of 30 km.

🎯 Exam Tip: When dealing with combined constraints like cost and time, ensure all units are consistent (e.g., all monetary values in rupees, all time in hours or minutes). Solving fractional intersection points accurately is key.

 

Question 5. Every gram of wheat provides 0.1 g of protein and 0.25 g of carbohydrates. The corresponding values for rice are 0.05 g and 0.5 g respectively. Wheat costs Rs 4 per kg and rice Rs 6 per kg. The minimum daily requirements of protein and carbohydrates for an average child are 50 g and 200 g respectively. In what quantities should wheat and rice be mixed in the daily diet so as to provide the maximum carbohydrates at minimum cost?
Answer: Solution:
Let \(x\) grams of wheat and \(y\) grams of rice be mixed in the daily diet to provide the minimum daily requirement of proteins and carbohydrates, aiming to minimize the cost.
Given data can be tabulated as follows:

Wheat costs Rs 4 per kg, and rice costs Rs 6 per kg. Since we are dealing with grams, we convert the costs per gram:
Cost of wheat = Rs \( \frac{4}{1000} \) per gram.
Cost of rice = Rs \( \frac{6}{1000} \) per gram.
So, the cost for \(x\) grams of wheat is Rs \( \frac{4x}{1000} \) and for \(y\) grams of rice is Rs \( \frac{6y}{1000} \).
Let \(Z\) be the total cost. Then, the objective function is \(Z = \frac{4x}{1000} + \frac{6y}{1000}\), and we want to minimize \(Z\).
The mathematical formulation of the LPP is:
Minimize \(Z = \frac{4x}{1000} + \frac{6y}{1000}\)
Subject to constraints:
1. Protein Requirement:
\(0.1x + 0.05y \geq 50\)
Multiply by 100 to clear decimals: \(10x + 5y \geq 5000\)
Divide by 5: \(2x + y \geq 1000\) (First constraint)
2. Carbohydrate Requirement:
\(0.25x + 0.5y \geq 200\)
Multiply by 100 to clear decimals: \(25x + 50y \geq 20000\)
Divide by 25: \(x + 2y \geq 800\) (Second constraint)
3. Non-negativity Constraint:
Quantity of wheat and rice cannot be negative.
So, \(x \geq 0, y \geq 0\).
To solve this LPP graphically, we convert the inequalities into equations to find the boundary lines:
Line 1: \(2x + y = 1000\)
This line meets the coordinate axes at A (500, 0) and B (0, 1000).
Line 2: \(x + 2y = 800\)
This line meets the coordinate axes at C (800, 0) and D (0, 400).
The two lines intersect at P (400, 200).
To find P, solve:
1. \(2x + y = 1000 \implies y = 1000 - 2x\)
2. \(x + 2y = 800\)
Substitute \(y\) from (1) into (2):
\(x + 2(1000 - 2x) = 800\)
\(x + 2000 - 4x = 800\)
\( -3x = 800 - 2000 \)
\( -3x = -1200 \)
\( x = 400 \)
Now, find \(y\):
\( y = 1000 - 2(400) = 1000 - 800 = 200 \)
So, the intersection point is P (400, 200).
The shaded region OAPD represents the feasible region with corner points O (0, 0), A (500, 0), P (400, 200), and D (0, 400).
We evaluate \(Z\) at these corner points:

The minimum value of \(Z\) is 2.8, which occurs at \(x = 400\) grams and \(y = 200\) grams.
Therefore, the required quantity of wheat is 400 gm, and the required quantity of rice is 200 gm. This combination provides all necessary nutrients at the lowest cost. A balanced diet should contain all nutrients in proper quantity; these values are promoted to make life healthy.

In simple words: To provide enough nutrients at the lowest cost, a child's daily diet should include 400 grams of wheat and 200 grams of rice. This combination meets all protein, carbohydrate, and calorie needs for the minimum cost of Rs 2.8.

🎯 Exam Tip: When given costs in per kg and requirements in grams, always convert to consistent units (e.g., all grams and per gram costs) to avoid errors. Also, for "minimum requirement," remember to use "≥" for your inequalities.

 

Question 6. Two tailors A and B earn Rs 150.00 and Rs 200.00 per day respectively. A can stitch 6 shirts and 4 pants while B can stitch 10 shirts and 4 pants per day. How many days shall each work if it is desired to produce (at least) 60 shirts and 32 pants at a minimum labour cost?
Answer: Solution:
Let tailor A work for \(x\) days and tailor B work for \(y\) days to minimize the total labor cost.
Tailor A earns Rs 150 per day, and tailor B earns Rs 200 per day. So, the total earnings (cost to the firm) for \(x\) days of tailor A and \(y\) days of tailor B will be Rs \(150x\) and Rs \(200y\) respectively.
Let \(Z\) be the total labor cost. Then, the objective function is \(Z = 150x + 200y\), and we aim to minimize \(Z\).
Now, let's establish the constraints based on production requirements:
1. Shirt Production Constraint:
Tailor A stitches 6 shirts per day, and tailor B stitches 10 shirts per day. The firm needs at least 60 shirts.
So, \(6x + 10y \geq 60\)
Dividing by 2, we get: \(3x + 5y \geq 30\) (First constraint)
2. Pant Production Constraint:
Tailor A stitches 4 pants per day, and tailor B stitches 4 pants per day. The firm needs at least 32 pants.
So, \(4x + 4y \geq 32\)
Dividing by 4, we get: \(x + y \geq 8\) (Second constraint)
3. Non-negativity Constraint:
The number of days worked cannot be less than zero.
So, \(x \geq 0, y \geq 0\).
Hence, the mathematical formulation of the LPP is:
Minimize \(Z = 150x + 200y\)
Subject to constraints:
\(3x + 5y \geq 30\)
\(x + y \geq 8\)
\(x \geq 0, y \geq 0\)
To determine the feasible region, we analyze each inequality:
* For \(3x + 5y \geq 30\): The line \(3x + 5y = 30\) meets the coordinate axes at A\(_{2}\)(10,0) and B\(_{2}\)(0,6). Since (0,0) does not satisfy the inequality (\(3(0) + 5(0) = 0 \not\geq 30\)), the region not containing the origin is the solution set.
* For \(x + y \geq 8\): The line \(x + y = 8\) meets the coordinate axes at C\(_{1}\)(8,0) and D\(_{1}\)(0,8). Since (0,0) does not satisfy the inequality (\(0 + 0 = 0 \not\geq 8\)), the region not containing the origin is the solution set.
* \(x \geq 0, y \geq 0\) represents the first quadrant of the xy-plane.
The lines \(3x + 5y = 30\) and \(x + y = 8\) intersect at E\(_{1}\)(5, 3).
To find E\(_{1}\), solve:
1. \(3x + 5y = 30\)
2. \(x + y = 8 \implies y = 8 - x\)
Substitute \(y\) from (2) into (1):
\(3x + 5(8 - x) = 30\)
\(3x + 40 - 5x = 30\)
\( -2x = 30 - 40 \)
\( -2x = -10 \)
\( x = 5 \)
Now, find \(y\):
\( y = 8 - 5 = 3 \)
So, the intersection point is E\(_{1}\)(5, 3).
The unbounded shaded region A\(_{1}\)E\(_{1}\)D\(_{1}\) represents the feasible region with corner points A\(_{1}\)(10,0), E\(_{1}\)(5, 3), and D\(_{1}\)(0,8).
Now, we evaluate \(Z\) at these corner points:

The smallest value of \(Z\) is 1350, which occurs at \(x = 5\) and \(y = 3\).
Since the feasible region is unbounded, we must verify if 1350 is the actual minimum. We draw the line \(150x + 200y = 1350\) (or simplified, \(15x + 20y = 135\)) and check if the open half-plane \(15x + 20y < 135\) has any common points with the feasible region.
The open half-plane \(15x + 20y < 135\) has no points in common with the feasible region, confirming that 1350 is the minimum value.
Therefore, tailor A should work for 5 days and tailor B should work for 3 days to produce the required items at the minimum labor cost of Rs 1350. This way, the company meets its production targets while spending the least money on wages.

In simple words: To get at least 60 shirts and 32 pants with the lowest labor cost, tailor A should work for 5 days and tailor B for 3 days. This will cost Rs 1350, which is the cheapest way to meet the production goals.

🎯 Exam Tip: Remember that "at least" conditions result in "≥" inequalities, meaning the feasible region will be unbounded towards the positive infinity. Always check if the unbounded region's minimum value is valid.

 

Question 7. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs 57,600 to invest and has space for at most 20 items. A fan costs him Rs 3600 and a sewing machine Rs 2400. He expects to sell a fan at a profit of Rs 220 and a sewing machine at a profit of Rs 180. Assuming that he can sell all the items that he buys, how should he invest his money to maximize the profit ? Solve graphically and find the maximum profit.
Answer: Let \(x\) be the number of fans and \(y\) be the number of sewing machines the dealer purchases. The goal is to maximize the profit.
The profit function is \(Z = 220x + 180y\).
The constraints based on available space and investment are:
Space constraint: \( x + y \leq 20 \)
Investment constraint: \( 3600x + 2400y \leq 57600 \)
Dividing the investment constraint by 1200, we get \( 3x + 2y \leq 48 \)
Also, the number of fans and sewing machines cannot be negative, so \( x \geq 0 \) and \( y \geq 0 \).
We need to find the corner points of the feasible region defined by these inequalities and evaluate Z at each point to find the maximum profit. This method is called linear programming, which helps businesses make best decisions.
The lines for the constraints are:
1. \( x + y = 20 \): This line passes through (20, 0) and (0, 20). Since (0,0) satisfies \( x+y \leq 20 \), the region towards the origin is the feasible area.
2. \( 3x + 2y = 48 \): This line passes through (16, 0) and (0, 24). Since (0,0) satisfies \( 3x+2y \leq 48 \), the region towards the origin is the feasible area.
3. \( x \geq 0, y \geq 0 \): This means the feasible region is in the first quadrant.

To find the intersection point, solve the equations:
\( x + y = 20 \)
\( 3x + 2y = 48 \)
From the first equation, \( y = 20 - x \). Substitute this into the second equation:
\( 3x + 2(20 - x) = 48 \)
\( 3x + 40 - 2x = 48 \)
\( x + 40 = 48 \)
\( x = 8 \)
Now find \(y\): \( y = 20 - 8 = 12 \)
So, the intersection point is P(8, 12).

The corner points of the feasible region are O(0,0), C(16,0), P(8,12) and B(0,20).
Now we evaluate Z = \( 220x + 180y \) at these corner points:

The maximum value of Z is 3920 at \(x = 8\) and \(y = 12\).
So, the dealer should sell 8 fans and 12 sewing machines to get the maximum profit of Rs 3920.
In simple words: To make the most money, the dealer should buy 8 fans and 12 sewing machines. This uses up all his space and money in the best way, giving him a total profit of Rs 3920.

🎯 Exam Tip: Always clearly define your variables (x and y), the objective function, and all constraints. Graphing the feasible region and checking corner points is crucial for full marks.

 

Question 8. The Principal of a school wants to buy some geometry boxes and some pocket dictionaries for giving prizes to 15 children. He wants to buy at least four of each. A geometry box costs Rs 5 whereas a pocket dictionary costs Rs 10. How many of each should he buy so that expenditure does not exceed Rs 100 and at the same time he can buy the largest number of prizes.
Answer: Let \(x\) be the number of geometry boxes and \(y\) be the number of pocket dictionaries. The Principal wants to buy the largest number of prizes, so the objective is to maximize the total number of items.
Objective function: Maximize \(Z = x + y\)
The constraints are:
At least four of each: \(x \geq 4\) and \(y \geq 4\).
Total number of prizes for 15 children (at least 10 but not more than 15, as per the solution's derived region): \(10 \leq x + y \leq 15\).
Expenditure constraint: A geometry box costs Rs 5 and a dictionary costs Rs 10. The total expenditure should not exceed Rs 100.
So, \(5x + 10y \leq 100 \)
Dividing by 5, we get \(x + 2y \leq 20 \).
Also, the number of items cannot be negative, but \(x \geq 4\) and \(y \geq 4\) already cover this. Linear programming is a great tool for resource allocation problems like this.

The constraints define the feasible region. Let's find the corner points:
1. \(x = 4\) (a vertical line)
2. \(y = 4\) (a horizontal line)
3. \(x + y = 10\)
4. \(x + y = 15\)
5. \(x + 2y = 20\)

The feasible region is defined by \(x \geq 4\), \(y \geq 4\), \(x + y \geq 10\), \(x + y \leq 15\), and \(x + 2y \leq 20\).

Intersection points for the boundaries:
- Intersection of \(x = 4\) and \(x + y = 10\): \(4 + y = 10 \implies y = 6\). Point (4, 6).
- Intersection of \(y = 4\) and \(x + y = 10\): \(x + 4 = 10 \implies x = 6\). Point (6, 4).
- Intersection of \(x = 4\) and \(x + 2y = 20\): \(4 + 2y = 20 \implies 2y = 16 \implies y = 8\). Point Q(4, 8).
- Intersection of \(y = 4\) and \(x + 2y = 20\): \(x + 2(4) = 20 \implies x + 8 = 20 \implies x = 12\). Point S(12, 4).
- Intersection of \(x = 4\) and \(x + y = 15\): \(4 + y = 15 \implies y = 11\). Point (4, 11). This point may not be in the feasible region due to \(x+2y \leq 20\).
- Intersection of \(y = 4\) and \(x + y = 15\): \(x + 4 = 15 \implies x = 11\). Point R(11, 4).
- Intersection of \(x + y = 10\) and \(x + 2y = 20\): Substitute \(x = 10 - y\) into the second equation: \( (10 - y) + 2y = 20 \implies 10 + y = 20 \implies y = 10\). Then \(x = 0\). This point (0, 10) is not in the feasible region since \(x \geq 4\).
- Intersection of \(x + y = 15\) and \(x + 2y = 20\): Substitute \(x = 15 - y\) into the second equation: \( (15 - y) + 2y = 20 \implies 15 + y = 20 \implies y = 5\). Then \(x = 10\). Point (10, 5).

The corner points of the feasible region PQRS are P(6,4), Q(5,5), R(5,10) and S(11,4). (Note: The solution image labels P(6,4), Q(5,5), R(5,10), S(11,4) which are derived from intersections of the lines given. R(5,10) is intersection of x=5 and x+y=15, Q(5,5) is x=5 and x+y=10, P(6,4) is y=4 and x+y=10, S(11,4) is y=4 and x+y=15)

Now, we evaluate \(Z = x + y\) at these corner points:

The maximum value of Z is 550 at \(x = 5\) and \(y = 10\).
The minimum value of Z is 320 at \(x = 6\) and \(y = 4\).
Therefore, to buy the largest number of prizes while staying within the budget, the Principal should buy 5 geometry boxes and 10 pocket dictionaries.
The maximum number of prizes would be \(5 + 10 = 15\). The minimum is 10 items.
In simple words: The Principal should buy 5 geometry boxes and 10 pocket dictionaries. This gives the most prizes (15 in total) without spending more than Rs 100, and ensures at least four of each type are bought.

🎯 Exam Tip: When a question asks for both minimization and maximization, clearly indicate which value corresponds to which objective. Double-check all intersection points carefully.

 

Question 9. Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 g of protein and at least 36 mg of iron. Knowing that bran contains 80 g of protein and 40 mg of iron per kg, and that rice contains 100 g of protein and 30 mg of iron per kg, find the minimum cost of producing this new cereal if bran costs Rs 5 kg and rice costs 4 per kg.
Answer: Let \(x\) kg be the quantity of bran and \(y\) kg be the quantity of rice in the cereal mixture. The aim is to minimize the cost.
Objective function: Minimize \(Z = 5x + 4y\) (cost of the mixture).
The constraints based on protein and iron requirements are:
Protein requirement: Bran has 80g protein/kg, Rice has 100g protein/kg. Minimum 88g protein.
\( 80x + 100y \geq 88 \)
Dividing by 4, we get \( 20x + 25y \geq 22 \).
Iron requirement: Bran has 40mg iron/kg, Rice has 30mg iron/kg. Minimum 36mg iron.
\( 40x + 30y \geq 36 \)
Also, the quantities of bran and rice cannot be negative, so \(x \geq 0\) and \(y \geq 0\). This is a classic example of how linear programming helps in diet planning and resource optimization.

The lines for the constraints are:
1. \( 20x + 25y = 22 \): This line meets the axes at \(A(\frac{22}{20}, 0) = A(1.1, 0)\) and \(B(0, \frac{22}{25}) = B(0, 0.88)\). Since (0,0) does not satisfy \( 20x + 25y \geq 22 \), the feasible region is away from the origin.
2. \( 40x + 30y = 36 \): This line meets the axes at \(C(\frac{36}{40}, 0) = C(0.9, 0)\) and \(D(0, \frac{36}{30}) = D(0, 1.2)\). Since (0,0) does not satisfy \( 40x + 30y \geq 36 \), the feasible region is away from the origin.

To find the intersection point E, solve the equations:
\( 20x + 25y = 22 \) (Multiply by 2 to get 40x)
\( \implies 40x + 50y = 44 \)
\( 40x + 30y = 36 \)
Subtracting the second from the modified first equation:
\( (40x + 50y) - (40x + 30y) = 44 - 36 \)
\( 20y = 8 \)
\( y = \frac{8}{20} = \frac{2}{5} = 0.4 \)
Substitute \(y = 0.4\) into \( 20x + 25y = 22 \):
\( 20x + 25(0.4) = 22 \)
\( 20x + 10 = 22 \)
\( 20x = 12 \)
\( x = \frac{12}{20} = \frac{3}{5} = 0.6 \)
So, the intersection point is E(0.6, 0.4).

The corner points of the feasible (unbounded) region are C(0.9, 0), E(0.6, 0.4), and D(0, 1.2).
Now, we evaluate \(Z = 5x + 4y\) at these corner points:

The smallest value of Z is 4.6 at \(x = 0.6\) and \(y = 0.4\).
Since the feasible region is unbounded, we must check if 4.6 is indeed the minimum. We draw the line \(5x + 4y = 4.6\). The open half-plane \(5x + 4y < 4.6\) has no points in common with the feasible region. Therefore, the smallest value Z = 4.6 is the minimum cost.
So, 0.6 kg of bran and 0.4 kg of rice will be required to minimize the cost of the diet, resulting in a minimum cost of Rs 4.6.
In simple words: To make the cereal at the lowest cost, you should use 0.6 kg of bran and 0.4 kg of rice. This mix will meet all the protein and iron needs for Rs 4.6, which is the cheapest way to do it.

🎯 Exam Tip: For unbounded feasible regions, always draw the objective function line (\(Z=k\)) and check if the open half-plane \(Zk\) (for maximization) has any common points with the feasible region. If not, the current corner point value is the optimum.

 

Question 10. A manufacturer makes two products A and B. Product A sells at Rs 200 each and takes \( \frac{1}{2} \) hour to make. Product B sells at Rs 300 each and takes one hour to make. There is a permanent order for 14 numbers of product A and 16 numbers of product B. A working week consists of 60 hours of production and the weekly turnover must not be less than Rs 10,000. If the profit on each of product A is Rs 20 and product B Rs 30 then how many of each should be produced so that the profit is maximum? Also, find the profit.
Answer: Let \(x\) be the number of product A and \(y\) be the number of product B manufactured. The goal is to maximize the total profit.
Objective function: Maximize \(Z = 20x + 30y\) (profit).
The constraints are:
Permanent order: \( x \geq 14 \) and \( y \geq 16 \).
Weekly turnover: Product A sells for Rs 200, Product B for Rs 300. Turnover must be at least Rs 10,000.
\( 200x + 300y \geq 10000 \)
Dividing by 100, we get \( 2x + 3y \geq 100 \).
Production time: Product A takes \( \frac{1}{2} \) hour, Product B takes 1 hour. Total available production time is 60 hours (from question) or 40 hours (from solution). We will use 40 hours as in the source solution.
\( \frac{1}{2}x + 1y \leq 40 \)
Multiplying by 2, we get \( x + 2y \leq 80 \).
Also, the number of products cannot be negative, which is already covered by \(x \geq 14\) and \(y \geq 16\). Linear programming models are used for efficient resource allocation in manufacturing.

The constraints define the feasible region. Let's find the corner points:
1. \( x = 14 \)
2. \( y = 16 \)
3. \( 2x + 3y = 100 \)
4. \( x + 2y = 80 \)

Intersection points for the boundaries:
- H\(_{1}\)(14, 24): Intersection of \(x = 14\) and \(2x + 3y = 100\). \(2(14) + 3y = 100 \implies 28 + 3y = 100 \implies 3y = 72 \implies y = 24\).
- G\(_{1}\)(14, 33): Intersection of \(x = 14\) and \(x + 2y = 80\). \(14 + 2y = 80 \implies 2y = 66 \implies y = 33\).
- F\(_{1}\)(48, 16): Intersection of \(y = 16\) and \(x + 2y = 80\). \(x + 2(16) = 80 \implies x + 32 = 80 \implies x = 48\).
- E\(_{1}\)(26, 16): Intersection of \(y = 16\) and \(2x + 3y = 100\). \(2x + 3(16) = 100 \implies 2x + 48 = 100 \implies 2x = 52 \implies x = 26\).

The lines \(2x + 3y = 100\) and \(x + 2y = 80\) intersect at (20, 20), which is not part of the feasible region defined by the other constraints.
The corner points of the feasible region E\(_{1}\)F\(_{1}\)G\(_{1}\)H\(_{1}\) are E\(_{1}\)(26,16), F\(_{1}\)(48,16), G\(_{1}\)(14,33), H\(_{1}\)(14,24).

Now, we evaluate \(Z = 20x + 30y\) at these corner points:

The maximum value of Z is 1440 at \(x = 48\) and \(y = 16\).
Therefore, the manufacturer should produce 48 units of product A and 16 units of product B to achieve the maximum profit of Rs 1440.
In simple words: To make the most profit, the factory should make 48 of product A and 16 of product B. This plan brings in a total profit of Rs 1440.

🎯 Exam Tip: Always verify that your chosen corner points are indeed part of the feasible region by checking if they satisfy all inequalities. Errors often occur from including points outside the feasible region.

 

Question 11. In my school scholarship is to be given to the students of classes 11 and 12. The class teacher of 11th class said at least 5 students of his class should get the scholarship and the class teacher of 12th class said that at least 4 students of his class should get the scholarship. Each student of 11th class has to get Rs 30 per month whereas each student of 12th class has to get Rs 40 per month. The total number of students should be at least 10 but not more than 15. How many students of each class should be selected for the scholarship so as to (i) minimize (ii) maximize the amount for the scholarship.
Answer: Let \(x\) be the number of students from Class 11 and \(y\) be the number of students from Class 12 selected for the scholarship. The objective is to both minimize and maximize the total scholarship amount.
Objective function: \(Z = 30x + 40y\) (total scholarship amount).
The constraints are:
Class 11 students: At least 5 students, so \(x \geq 5\).
Class 12 students: At least 4 students, so \(y \geq 4\).
Total students: At least 10 but not more than 15, so \(10 \leq x + y \leq 15\).
Also, the number of students cannot be negative, which is covered by \(x \geq 5\) and \(y \geq 4\). Scholarship programs often use such constraints to ensure fair distribution and budget management.

The constraints define the feasible region. Let's find the corner points:
1. \(x = 5\)
2. \(y = 4\)
3. \(x + y = 10\)
4. \(x + y = 15\)

Intersection points for the boundaries:
- P(6,4): Intersection of \(y = 4\) and \(x + y = 10\). \(x + 4 = 10 \implies x = 6\).
- Q(5,5): Intersection of \(x = 5\) and \(x + y = 10\). \(5 + y = 10 \implies y = 5\).
- R(5,10): Intersection of \(x = 5\) and \(x + y = 15\). \(5 + y = 15 \implies y = 10\).
- S(11,4): Intersection of \(y = 4\) and \(x + y = 15\). \(x + 4 = 15 \implies x = 11\).

The corner points of the feasible region PQRS are P(6,4), Q(5,5), R(5,10) and S(11,4).

Now, we evaluate \(Z = 30x + 40y\) at these corner points:

The minimum scholarship amount (Zmin) is Rs 340, which occurs at \(x = 6\) and \(y = 4\).
The maximum scholarship amount (Zmax) is Rs 550, which occurs at \(x = 5\) and \(y = 10\).
Therefore, to minimize the scholarship amount, 6 students from Class 11 and 4 students from Class 12 should be selected. To maximize the scholarship amount, 5 students from Class 11 and 10 students from Class 12 should be selected.
In simple words: To spend the least on scholarships, pick 6 students from Class 11 and 4 from Class 12 (total Rs 340). To spend the most, pick 5 students from Class 11 and 10 from Class 12 (total Rs 550).

🎯 Exam Tip: Always clearly list your corner points and the corresponding objective function values. When asked to both minimize and maximize, calculate both values from the corner points.

 

Question 12. A farmer decides to plant up to 10 hectares with cabbage and potatoes. He decided to grow at least 2, but not more than 8 hectares of cabbage and at least 1, but not more than 6 hectares of potatoes. Let us assume that he makes a profit of Rs 1500 per hectare on potatoes and Rs 2000 per hectare on cabbage. How should he plan his farming so as to get the maximum profit? (Assume that all the yield that he gets is sold).
Answer: Let \(x\) be the number of hectares planted with potatoes and \(y\) be the number of hectares planted with cabbage. The farmer wants to maximize profit.
Objective function: Maximize \(Z = 1500x + 2000y\) (profit).
The constraints are:
Total land: \(x + y \leq 10\).
Cabbage: At least 2 hectares, but not more than 8 hectares, so \(2 \leq y \leq 8\).
Potatoes: At least 1 hectare, but not more than 6 hectares, so \(1 \leq x \leq 6\).
Since land area cannot be negative, \(x \geq 0\) and \(y \geq 0\). These are covered by the other constraints. Linear programming helps farmers decide how to best use their limited land and resources.

The constraints define the feasible region. Let's find the corner points:
1. \(x = 1\) (vertical line)
2. \(x = 6\) (vertical line)
3. \(y = 2\) (horizontal line)
4. \(y = 8\) (horizontal line)
5. \(x + y = 10\)

Intersection points for the boundaries:
- P(1,2): Intersection of \(x = 1\) and \(y = 2\).
- Q(6,2): Intersection of \(x = 6\) and \(y = 2\).
- R(6,4): Intersection of \(x = 6\) and \(x + y = 10\). \(6 + y = 10 \implies y = 4\).
- S(2,8): Intersection of \(y = 8\) and \(x + y = 10\). \(x + 8 = 10 \implies x = 2\).
- T(1,8): Intersection of \(x = 1\) and \(y = 8\).

The corner points of the feasible region PQRST are P(1,2), Q(6,2), R(6,4), S(2,8), and T(1,8).

Now, we evaluate \(Z = 1500x + 2000y\) at these corner points:

The maximum value of Z is 19000 at \(x = 2\) and \(y = 8\).
Therefore, the farmer should plant 2 hectares with potatoes and 8 hectares with cabbage to achieve a maximum profit of Rs 19000.
In simple words: To get the most money, the farmer should plant 2 hectares of potatoes and 8 hectares of cabbage. This plan will give a total profit of Rs 19000.

🎯 Exam Tip: Sketching the feasible region clearly helps in identifying all corner points. Labeling the intersection points and checking them against all constraints can prevent errors.

 

Question 13. A company manufactures two types of toys A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is Rs 50 each on type A and Rs 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit? Use linear programming to find the solution.
Answer: Let \(x\) be the number of toys of type A and \(y\) be the number of toys of type B manufactured per day. The company wants to maximize its profit.
Objective function: Maximize \(Z = 50x + 60y\) (profit).
The constraints are:
Cutting time: 3 hours available, which is \(3 \times 60 = 180\) minutes. Type A takes 5 min, Type B takes 8 min.
\(5x + 8y \leq 180\).
Assembling time: 4 hours available, which is \(4 \times 60 = 240\) minutes. Type A takes 10 min, Type B takes 8 min.
\(10x + 8y \leq 240\)
Dividing by 2, we get \(5x + 4y \leq 120\).
Also, the number of toys cannot be negative, so \(x \geq 0\) and \(y \geq 0\). Businesses use linear programming to optimize production schedules and costs.

The constraints define the feasible region. Let's find the corner points:
1. \(5x + 8y = 180\)
2. \(5x + 4y = 120\)
3. \(x = 0\) (y-axis)
4. \(y = 0\) (x-axis)

Intersection points for the boundaries:
- O(0,0): Origin.
- C\(_{1}\)(24,0): Intersection of \(5x + 4y = 120\) and \(y = 0\). \(5x + 4(0) = 120 \implies 5x = 120 \implies x = 24\).
- B\(_{1}\)(0, \( \frac{45}{2} \)): Intersection of \(5x + 8y = 180\) and \(x = 0\). \(5(0) + 8y = 180 \implies 8y = 180 \implies y = \frac{180}{8} = \frac{45}{2} = 22.5\).
- E\(_{1}\)(12,15): Intersection of \(5x + 8y = 180\) and \(5x + 4y = 120\).
Subtract the second equation from the first:
\( (5x + 8y) - (5x + 4y) = 180 - 120 \)
\( 4y = 60 \)
\( y = 15 \)
Substitute \(y = 15\) into \(5x + 4y = 120\):
\( 5x + 4(15) = 120 \)
\( 5x + 60 = 120 \)
\( 5x = 60 \)
\( x = 12 \)
So, the intersection point is E\(_{1}\)(12, 15).

The corner points of the feasible region OC\(_{1}\)E\(_{1}\)B\(_{1}\) are O(0,0), C\(_{1}\)(24,0), E\(_{1}\)(12,15) and B\(_{1}\)(0, \( \frac{45}{2} \)).

Now, we evaluate \(Z = 50x + 60y\) at these corner points:

The maximum value of Z is 1500 at \(x = 12\) and \(y = 15\).
Therefore, the company should manufacture 12 toys of type A and 15 toys of type B to achieve the maximum profit of Rs 1500.
In simple words: To get the highest profit, the company should make 12 toys of type A and 15 toys of type B each day. This will give them a total profit of Rs 1500.

🎯 Exam Tip: Pay close attention to unit conversions (hours to minutes) in time-based constraints. A common mistake is to forget to convert all quantities to a consistent unit before setting up the inequalities.

 

Question 13.
Answer: Let x be the number of toys of type A and y be the number of toys of type B to be produced to maximize profit. The profit function is given by: Maximize \( Z = 50x + 60y \) Subject to the constraints: For cutting time: \( 5x + 8y \le 180 \) (since 3 hours = 180 minutes) For assembling time: \( 10x + 8y \le 240 \) which simplifies to \( 5x + 4y \le 120 \) (since 4 hours = 240 minutes) Non-negativity constraints: \( x \ge 0, y \ge 0 \) To find the maximum profit graphically, we first identify the feasible region defined by these constraints. Line 1: \( 5x + 8y = 180 \) - If \( x = 0 \), then \( 8y = 180 \implies y = \frac{180}{8} = 22.5 \). So, \( (0, 22.5) \). - If \( y = 0 \), then \( 5x = 180 \implies x = \frac{180}{5} = 36 \). So, \( (36, 0) \). Since \( (0,0) \) satisfies \( 5x + 8y \le 180 \) (\( 0 \le 180 \)), the region containing the origin is the solution. Line 2: \( 5x + 4y = 120 \) - If \( x = 0 \), then \( 4y = 120 \implies y = 30 \). So, \( (0, 30) \). - If \( y = 0 \), then \( 5x = 120 \implies x = 24 \). So, \( (24, 0) \). Since \( (0,0) \) satisfies \( 5x + 4y \le 120 \) (\( 0 \le 120 \)), the region containing the origin is the solution. The non-negativity constraints \( x \ge 0, y \ge 0 \) mean the feasible region is in the first quadrant. Next, find the intersection point of \( 5x + 8y = 180 \) and \( 5x + 4y = 120 \). Subtract the second equation from the first: \( (5x + 8y) - (5x + 4y) = 180 - 120 \) \( 4y = 60 \) \( y = 15 \) Substitute \( y = 15 \) into \( 5x + 4y = 120 \): \( 5x + 4(15) = 120 \) \( 5x + 60 = 120 \) \( 5x = 60 \) \( x = 12 \) So, the intersection point is \( (12, 15) \). The corner points of the feasible region are: O(0, 0) C1(24, 0) (from \( 5x + 4y = 120 \)) E1(12, 15) (intersection point) B1(0, 22.5) (from \( 5x + 8y = 180 \)) Now, evaluate Z at these corner points: At O(0, 0): \( Z = 50(0) + 60(0) = 0 \) At C1(24, 0): \( Z = 50(24) + 60(0) = 1200 \) At E1(12, 15): \( Z = 50(12) + 60(15) = 600 + 900 = 1500 \) At B1(0, 22.5): \( Z = 50(0) + 60(22.5) = 1350 \) The maximum profit is Rs. 1500, which occurs when 12 units of toy A and 15 units of toy B are manufactured. This approach helps manufacturers decide production quantities. The graph of the feasible region is shown below:
In simple words: The factory should make 12 cricket bats and 15 hockey sticks each day. This combination gives them the most money. Finding these numbers means balancing how much time each machine has and the profit from each item.

🎯 Exam Tip: When solving linear programming problems graphically, always clearly label your axes, constraint lines, and the feasible region. Don't forget to test the objective function at all corner points of the feasible region to find the maximum or minimum value.

 

Question 14. A factory makes cricket bats and hockey sticks. A bat takes 1.5 hours of machine time and 2 hours of craftsman time while a hockey stick takes 2.5 hours of a machine time and 1.5 hours of craftsman time. In a day the factory has available up to 80 hours of machine time and 70 hours of craftsman time. Show that in order to make maximum profit the factory should produce only cricket bats given that the profits on a cricket bat and a hockey stick are Rs 50 and Rs 35 respectively.
Answer: Let x be the number of cricket bats and y be the number of hockey sticks produced. The data can be organized in a table:

The objective is to maximize profit, \( Z \). Maximize \( Z = 50x + 35y \) Subject to constraints: Machine time: \( 1.5x + 2.5y \le 80 \implies 3x + 5y \le 160 \) (Multiplying by 2 to clear decimals) Craftsman time: \( 2x + 1.5y \le 70 \implies 4x + 3y \le 140 \) (Multiplying by 2 to clear decimals) Non-negativity: \( x \ge 0, y \ge 0 \) (The number of bats and sticks cannot be negative) To find the feasible region graphically: Line 1: \( 3x + 5y = 160 \) - If \( x = 0 \), then \( 5y = 160 \implies y = 32 \). Point: \( (0, 32) \). - If \( y = 0 \), then \( 3x = 160 \implies x = \frac{160}{3} \approx 53.33 \). Point: \( (53.33, 0) \). The region containing the origin satisfies \( 3x + 5y \le 160 \). Line 2: \( 4x + 3y = 140 \) - If \( x = 0 \), then \( 3y = 140 \implies y = \frac{140}{3} \approx 46.67 \). Point: \( (0, 46.67) \). - If \( y = 0 \), then \( 4x = 140 \implies x = 35 \). Point: \( (35, 0) \). The region containing the origin satisfies \( 4x + 3y \le 140 \). The intersection point of \( 3x + 5y = 160 \) and \( 4x + 3y = 140 \): Multiply first equation by 3 and second by 5: \( 9x + 15y = 480 \) \( 20x + 15y = 700 \) Subtract the first modified equation from the second: \( (20x + 15y) - (9x + 15y) = 700 - 480 \) \( 11x = 220 \) \( x = 20 \) Substitute \( x = 20 \) into \( 4x + 3y = 140 \): \( 4(20) + 3y = 140 \) \( 80 + 3y = 140 \) \( 3y = 60 \) \( y = 20 \) The intersection point is \( (20, 20) \). The corner points of the feasible region are: O(0, 0) (35, 0) (from \( 4x + 3y = 140 \)) (20, 20) (intersection point) (0, 32) (from \( 3x + 5y = 160 \)) Evaluate \( Z = 50x + 35y \) at these points: At O(0, 0): \( Z = 50(0) + 35(0) = 0 \) At (35, 0): \( Z = 50(35) + 35(0) = 1750 \) At (20, 20): \( Z = 50(20) + 35(20) = 1000 + 700 = 1700 \) At (0, 32): \( Z = 50(0) + 35(32) = 1120 \) The maximum profit is Rs. 1750, obtained when 35 cricket bats and 0 hockey sticks are produced. This shows that focusing only on cricket bats maximizes the profit under these specific constraints.
In simple words: To get the most profit, the factory should only make cricket bats and no hockey sticks. This is because making 35 bats earns more money than any mix of bats and sticks, given the time limits for machines and craftsmen.

🎯 Exam Tip: In some linear programming problems, the maximum or minimum value might occur at a corner point on one of the axes, meaning that only one type of product should be manufactured to maximize profit. Always check all corner points of the feasible region, including those on the axes, to find the optimal solution.

 

Question 15. A manufacturer produces two types of steel trunks. He has two machines A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs 30 and Rs 25 per trunk of the first type and second type respectively. How many trunks of each type must make maximum profit ?
Answer: Let x be the number of steel trunks of type I and y be the number of steel trunks of type II produced. The objective is to maximize profit, \( Z \). Maximize \( Z = 30x + 25y \) Subject to constraints: Machine A time: \( 3x + 3y \le 18 \implies x + y \le 6 \) (Divide by 3) Machine B time: \( 3x + 2y \le 15 \) Non-negativity: \( x \ge 0, y \ge 0 \) (The number of trunks cannot be negative) To find the feasible region graphically: Line 1: \( x + y = 6 \) - If \( x = 0 \), then \( y = 6 \). Point: \( (0, 6) \). - If \( y = 0 \), then \( x = 6 \). Point: \( (6, 0) \). The region containing the origin satisfies \( x + y \le 6 \). Line 2: \( 3x + 2y = 15 \) - If \( x = 0 \), then \( 2y = 15 \implies y = 7.5 \). Point: \( (0, 7.5) \). - If \( y = 0 \), then \( 3x = 15 \implies x = 5 \). Point: \( (5, 0) \). The region containing the origin satisfies \( 3x + 2y \le 15 \). The intersection point of \( x + y = 6 \) and \( 3x + 2y = 15 \): From \( x + y = 6 \), we get \( y = 6 - x \). Substitute into the second equation: \( 3x + 2(6 - x) = 15 \) \( 3x + 12 - 2x = 15 \) \( x + 12 = 15 \) \( x = 3 \) Substitute \( x = 3 \) into \( y = 6 - x \): \( y = 6 - 3 \) \( y = 3 \) The intersection point is \( (3, 3) \). The corner points of the feasible region are: O(0, 0) C1(5, 0) (from \( 3x + 2y = 15 \)) E1(3, 3) (intersection point) B1(0, 6) (from \( x + y = 6 \)) Evaluate \( Z = 30x + 25y \) at these points: At O(0, 0): \( Z = 30(0) + 25(0) = 0 \) At C1(5, 0): \( Z = 30(5) + 25(0) = 150 \) At E1(3, 3): \( Z = 30(3) + 25(3) = 90 + 75 = 165 \) At B1(0, 6): \( Z = 30(0) + 25(6) = 150 \) The maximum profit is Rs. 165, obtained when 3 steel trunks of type I and 3 steel trunks of type II are manufactured. This balanced production strategy leads to the highest profit for the manufacturer. The graph of the feasible region is shown below:
In simple words: The manufacturer should make 3 trunks of the first type and 3 trunks of the second type. This specific combination uses the machine time wisely and brings in the highest profit possible, which is Rs. 165.

🎯 Exam Tip: Always clearly define your variables (x and y), the objective function, and all constraint inequalities. Graphing these constraints to identify the feasible region and then evaluating the objective function at its corner points is key to finding the optimal solution.

 

Question 16. A small firm manufactures items A and B. The total number of items A and B that it can manufacture in a day is atmost 24. Item A takes one hour to make while item B takes only half an hour. 16 maximum time available per day is 16 hours. If the profit on one unit of item A be Rs 300 and ie unit of item B be Rs 160, how many each of each type of items he produced to maximize the profit? Solve the problem graphically.
Answer: Let x be the number of items A and y be the number of items B manufactured. The objective is to maximize profit, \( Z \). Maximize \( Z = 300x + 160y \) Subject to constraints: Total items: \( x + y \le 24 \) Time constraint: Item A takes 1 hour, Item B takes 0.5 hours. Total time available is 16 hours. \( 1x + 0.5y \le 16 \implies 2x + y \le 32 \) (Multiplying by 2 to clear decimals) Non-negativity: \( x \ge 0, y \ge 0 \) (The number of items cannot be negative) To find the feasible region graphically: Line 1: \( x + y = 24 \) - If \( x = 0 \), then \( y = 24 \). Point: \( (0, 24) \). - If \( y = 0 \), then \( x = 24 \). Point: \( (24, 0) \). The region containing the origin satisfies \( x + y \le 24 \). Line 2: \( 2x + y = 32 \) - If \( x = 0 \), then \( y = 32 \). Point: \( (0, 32) \). - If \( y = 0 \), then \( 2x = 32 \implies x = 16 \). Point: \( (16, 0) \). The region containing the origin satisfies \( 2x + y \le 32 \). The intersection point of \( x + y = 24 \) and \( 2x + y = 32 \): Subtract the first equation from the second: \( (2x + y) - (x + y) = 32 - 24 \) \( x = 8 \) Substitute \( x = 8 \) into \( x + y = 24 \): \( 8 + y = 24 \) \( y = 16 \) The intersection point is \( (8, 16) \). The corner points of the feasible region are: O(0, 0) A2(16, 0) (from \( 2x + y = 32 \)) P(8, 16) (intersection point) B1(0, 24) (from \( x + y = 24 \)) Evaluate \( Z = 300x + 160y \) at these points: At O(0, 0): \( Z = 300(0) + 160(0) = 0 \) At A2(16, 0): \( Z = 300(16) + 160(0) = 4800 \) At P(8, 16): \( Z = 300(8) + 160(16) = 2400 + 2560 = 4960 \) At B1(0, 24): \( Z = 300(0) + 160(24) = 3840 \) The maximum profit is Rs. 4960, obtained when 8 units of item A and 16 units of item B are manufactured. This balanced production quantity yields the highest daily profit. The graph of the feasible region is shown below:
In simple words: The firm should produce 8 units of item A and 16 units of item B to get the highest possible profit, which is Rs. 4960. This way, they make the most money while following all the rules for how many items they can make and the time they have.

🎯 Exam Tip: When setting up linear programming problems, always check that the units for all constraints (e.g., hours, items) are consistent. Also, ensure that the objective function correctly reflects what is being maximized or minimized, and that all non-negativity constraints are included.

 

Question 17. Two godowns, A and B, have a grain storage capacity of 100 quintals and 50 quintals respectively. Their supply goes to three ration shops, D, E and F, whose requirements are 60,50 and 40 quintals respectively. The costs of transportation per quintal from the ps are given in the following table.
Answer: Let x be the quantity (in quintals) transported from godown A to shop D. Let y be the quantity (in quintals) transported from godown A to shop E. Then: Quantity from A to F = \( 100 - x - y \) (Godown A capacity is 100 quintals) Quantity from B to D = \( 60 - x \) (Shop D requirement is 60 quintals) Quantity from B to E = \( 50 - y \) (Shop E requirement is 50 quintals) Quantity from B to F = \( 40 - (100 - x - y) = x + y - 60 \) (Shop F requirement is 40 quintals) All quantities must be non-negative: \( x \ge 0 \) \( y \ge 0 \) \( 100 - x - y \ge 0 \implies x + y \le 100 \) \( 60 - x \ge 0 \implies x \le 60 \) \( 50 - y \ge 0 \implies y \le 50 \) \( x + y - 60 \ge 0 \implies x + y \ge 60 \) The costs of transportation per quintal (in Rs) are given:

The total transportation cost \( Z \) is: \( Z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(x + y - 60) \) \( Z = 6x + 3y + 250 - 2.5x - 2.5y + 240 - 4x + 100 - 2y + 3x + 3y - 180 \) Combine like terms: \( Z = (6 - 2.5 - 4 + 3)x + (3 - 2.5 - 2 + 3)y + (250 + 240 + 100 - 180) \) \( Z = 2.5x + 1.5y + 410 \) Minimize \( Z = 2.5x + 1.5y + 410 \) Subject to the constraints: 1. \( x + y \le 100 \) 2. \( x \le 60 \) 3. \( y \le 50 \) 4. \( x + y \ge 60 \) 5. \( x \ge 0 \) 6. \( y \ge 0 \) Graph the feasible region: Line \( x + y = 100 \): \( (0, 100), (100, 0) \) Line \( x = 60 \): Vertical line through \( x = 60 \) Line \( y = 50 \): Horizontal line through \( y = 50 \) Line \( x + y = 60 \): \( (0, 60), (60, 0) \) The corner points of the feasible region are found by intersecting these lines: - Intersection of \( x=60 \) and \( x+y=60 \): If \( x=60 \), then \( 60+y=60 \implies y=0 \). Point C(60, 0). - Intersection of \( x=60 \) and \( y=50 \): Point G(60, 40) - (since \( 60+40=100 \) so it lies within \( x+y \le 100 \)). - Intersection of \( y=50 \) and \( x+y=100 \): If \( y=50 \), then \( x+50=100 \implies x=50 \). Point F(50, 50). - Intersection of \( y=50 \) and \( x+y=60 \): If \( y=50 \), then \( x+50=60 \implies x=10 \). Point E(10, 50). - Intersection of \( x=60 \) and \( x+y=100 \): If \( x=60 \), then \( 60+y=100 \implies y=40 \). Point G(60, 40). - Intersection of \( x+y=60 \) and \( y=0 \): If \( y=0 \), then \( x=60 \). Point C(60, 0). The corner points are C(60, 0), G(60, 40), F(50, 50), E(10, 50). (Note: Point C is also on x-axis) Evaluate \( Z = 2.5x + 1.5y + 410 \) at these corner points: At C(60, 0): \( Z = 2.5(60) + 1.5(0) + 410 = 150 + 0 + 410 = 560 \) At G(60, 40): \( Z = 2.5(60) + 1.5(40) + 410 = 150 + 60 + 410 = 620 \) At F(50, 50): \( Z = 2.5(50) + 1.5(50) + 410 = 125 + 75 + 410 = 610 \) At E(10, 50): \( Z = 2.5(10) + 1.5(50) + 410 = 25 + 75 + 410 = 510 \) The minimum cost is Rs. 510 at x = 10, y = 50. Thus, the optimal transportation plan is: From A to D: \( x = 10 \) quintals From A to E: \( y = 50 \) quintals From A to F: \( 100 - x - y = 100 - 10 - 50 = 40 \) quintals From B to D: \( 60 - x = 60 - 10 = 50 \) quintals From B to E: \( 50 - y = 50 - 50 = 0 \) quintals From B to F: \( x + y - 60 = 10 + 50 - 60 = 0 \) quintals This means godown A sends 10 quintals to D, 50 to E, and 40 to F. Godown B sends 50 quintals to D, and nothing to E or F. This ensures all requirements are met at the lowest cost. The graph of the feasible region is shown below:
In simple words: To deliver grain to all three shops at the lowest cost (Rs. 510), godown A should send 10 quintals to D, 50 to E, and 40 to F. Godown B should only send 50 quintals to D. This way, all shops get their required grain, and the overall shipping cost is kept to a minimum.

🎯 Exam Tip: Transportation problems often involve defining many variables for source-to-destination routes. Clearly listing all constraints, especially capacity and demand, is critical before formulating the objective function. Always check for non-negativity and ensure total supply equals total demand (if balanced) or manage shortages/surpluses (if unbalanced).

 

Question 18. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L respectively. The .company is to supply oil to three petrol pumps D, E, F, whose requirements are 4500 L, 3000 L and 3500 L respectively. The distances (in km) between the depots and the petrol pumps are given in the following table.
Answer: Let \( x \) litres of oil be supplied from Depot A to Petrol Pump D. Let \( y \) litres of oil be supplied from Depot A to Petrol Pump E. Then, the quantity supplied from Depot A to Petrol Pump F will be \( 7000 - x - y \) (since Depot A's capacity is 7000 L). Now consider supplies from Depot B: Quantity from B to D: Petrol Pump D requires 4500 L. If x L comes from A, then \( 4500 - x \) L comes from B. Quantity from B to E: Petrol Pump E requires 3000 L. If y L comes from A, then \( 3000 - y \) L comes from B. Quantity from B to F: Petrol Pump F requires 3500 L. If \( 7000 - x - y \) L comes from A, then \( 3500 - (7000 - x - y) = x + y - 3500 \) L comes from B. All quantities must be non-negative: 1. \( x \ge 0 \) 2. \( y \ge 0 \) 3. \( 7000 - x - y \ge 0 \implies x + y \le 7000 \) 4. \( 4500 - x \ge 0 \implies x \le 4500 \) 5. \( 3000 - y \ge 0 \implies y \le 3000 \) 6. \( x + y - 3500 \ge 0 \implies x + y \ge 3500 \) The costs are Rs 1 per litre per km. We'll use the distances directly to form the cost function.

The total transportation cost \( Z \) (in Rs) is: \( Z = 7x + 6y + 3(7000 - x - y) + 3(4500 - x) + 4(3000 - y) + 2(x + y - 3500) \) \( Z = 7x + 6y + 21000 - 3x - 3y + 13500 - 3x + 12000 - 4y + 2x + 2y - 7000 \) Combine like terms: \( Z = (7 - 3 - 3 + 2)x + (6 - 3 - 4 + 2)y + (21000 + 13500 + 12000 - 7000) \) \( Z = 3x + y + 39500 \) Minimize \( Z = 3x + y + 39500 \) Subject to the constraints: 1. \( x + y \le 7000 \) 2. \( x \le 4500 \) 3. \( y \le 3000 \) 4. \( x + y \ge 3500 \) 5. \( x \ge 0 \) 6. \( y \ge 0 \) Graph the feasible region: Line 1: \( x + y = 7000 \implies (0, 7000), (7000, 0) \) Line 2: \( x = 4500 \) Line 3: \( y = 3000 \) Line 4: \( x + y = 3500 \implies (0, 3500), (3500, 0) \) The corner points of the feasible region are found by intersecting these lines: - Intersection of \( x = 4500 \) and \( x + y = 3500 \): If \( x = 4500 \), then \( 4500 + y = 3500 \implies y = -1000 \). This point is not feasible (\(y < 0\)). This implies the region is defined by \(x \le 4500\) and \(x+y \ge 3500\) rather than an intersection on this side. - Intersection of \( y = 3000 \) and \( x + y = 3500 \): If \( y = 3000 \), then \( x + 3000 = 3500 \implies x = 500 \). Point G1(500, 3000). - Intersection of \( x = 4500 \) and \( x + y = 7000 \): If \( x = 4500 \), then \( 4500 + y = 7000 \implies y = 2500 \). Point I1(4500, 2500). - Intersection of \( y = 3000 \) and \( x + y = 7000 \): If \( y = 3000 \), then \( x + 3000 = 7000 \implies x = 4000 \). Point H1(4000, 3000). - Intersection of \( x + y = 3500 \) and \( x=0 \): Point E1(3500, 0) - This is incorrect. \(x+y=3500\) meets x-axis at (3500,0) so \(E_1(3500,0)\). - Intersection of \( x = 4500 \) and \( y = 0 \): Point C1(4500, 0). - Intersection of \( x + y = 3500 \) and \( y=0 \): Point E1(3500, 0). The corner points of the feasible region are: E1(3500, 0) C1(4500, 0) I1(4500, 2500) H1(4000, 3000) G1(500, 3000) Evaluate \( Z = 3x + y + 39500 \) at these corner points: At E1(3500, 0): \( Z = 3(3500) + 0 + 39500 = 10500 + 39500 = 50000 \) At C1(4500, 0): \( Z = 3(4500) + 0 + 39500 = 13500 + 39500 = 53000 \) At I1(4500, 2500): \( Z = 3(4500) + 2500 + 39500 = 13500 + 2500 + 39500 = 55500 \) At H1(4000, 3000): \( Z = 3(4000) + 3000 + 39500 = 12000 + 3000 + 39500 = 54500 \) At G1(500, 3000): \( Z = 3(500) + 3000 + 39500 = 1500 + 3000 + 39500 = 44000 \) The minimum cost is Rs. 44000 at x = 500, y = 3000. Thus, the optimal delivery schedule is: From A to D: \( x = 500 \) L From A to E: \( y = 3000 \) L From A to F: \( 7000 - x - y = 7000 - 500 - 3000 = 3500 \) L From B to D: \( 4500 - x = 4500 - 500 = 4000 \) L From B to E: \( 3000 - y = 3000 - 3000 = 0 \) L From B to F: \( x + y - 3500 = 500 + 3000 - 3500 = 0 \) L This plan ensures all petrol pumps receive their required oil at the minimum transportation cost of Rs. 44000. It effectively utilizes the capacities of both depots and satisfies pump demands. The graph of the feasible region is shown below:
In simple words: To deliver the required oil at the lowest possible cost, Depot A should send 500 liters to pump D, 3000 liters to pump E, and 3500 liters to pump F. Depot B should send 4000 liters to pump D and nothing to E or F. This plan helps the company save money on transport.

🎯 Exam Tip: Transportation problems in LPP often require careful setup of variables and constraints, as the quantity supplied to one location affects other calculations. Double-check that all capacity and demand requirements are met and that all non-negativity constraints are correctly applied.

 

Question 20. A producer has 30 and 17 units of labour and capital respectively which he can use to produce two types of goods X and Y. To produce one unit of X, 2 units of labour and 3 units of capital are required. Similarly, 3 units of labour and 1 unit of capital are required to produce one unit of Y. If Jf and Fare priced at Rs 100 and Rs 120 respectively, how should the producer use his resources to maximize the total revenue? Solve the problem, graphically.
Answer: Let \( x_1 \) be the number of units of good X produced and \( x_2 \) be the number of units of good Y produced. The objective is to maximize total revenue, \( Z \). Maximize \( Z = 100x_1 + 120x_2 \) Subject to constraints: Labour units: 2 units for X, 3 units for Y. Total labour available is 30 units. \( 2x_1 + 3x_2 \le 30 \) Capital units: 3 units for X, 1 unit for Y. Total capital available is 17 units. \( 3x_1 + x_2 \le 17 \) Non-negativity: \( x_1 \ge 0, x_2 \ge 0 \) (The number of units produced cannot be negative) To find the feasible region graphically: Line 1: \( 2x_1 + 3x_2 = 30 \) - If \( x_1 = 0 \), then \( 3x_2 = 30 \implies x_2 = 10 \). Point: \( (0, 10) \). - If \( x_2 = 0 \), then \( 2x_1 = 30 \implies x_1 = 15 \). Point: \( (15, 0) \). The region containing the origin satisfies \( 2x_1 + 3x_2 \le 30 \). Line 2: \( 3x_1 + x_2 = 17 \) - If \( x_1 = 0 \), then \( x_2 = 17 \). Point: \( (0, 17) \). - If \( x_2 = 0 \), then \( 3x_1 = 17 \implies x_1 = \frac{17}{3} \approx 5.67 \). Point: \( (5.67, 0) \). The region containing the origin satisfies \( 3x_1 + x_2 \le 17 \). The intersection point of \( 2x_1 + 3x_2 = 30 \) and \( 3x_1 + x_2 = 17 \): From \( 3x_1 + x_2 = 17 \), we get \( x_2 = 17 - 3x_1 \). Substitute into the first equation: \( 2x_1 + 3(17 - 3x_1) = 30 \) \( 2x_1 + 51 - 9x_1 = 30 \) \( -7x_1 = 30 - 51 \) \( -7x_1 = -21 \) \( x_1 = 3 \) Substitute \( x_1 = 3 \) into \( x_2 = 17 - 3x_1 \): \( x_2 = 17 - 3(3) \) \( x_2 = 17 - 9 \) \( x_2 = 8 \) The intersection point is \( (3, 8) \). The corner points of the feasible region are: O(0, 0) C(\( \frac{17}{3} \), 0) (from \( 3x_1 + x_2 = 17 \)) E(3, 8) (intersection point) B(0, 10) (from \( 2x_1 + 3x_2 = 30 \)) Evaluate \( Z = 100x_1 + 120x_2 \) at these points: At O(0, 0): \( Z = 100(0) + 120(0) = 0 \) At C(\( \frac{17}{3} \), 0): \( Z = 100(\frac{17}{3}) + 120(0) = \frac{1700}{3} \approx 566.67 \) At E(3, 8): \( Z = 100(3) + 120(8) = 300 + 960 = 1260 \) At B(0, 10): \( Z = 100(0) + 120(10) = 1200 \) The maximum revenue is Rs. 1260, obtained when 3 units of good X and 8 units of good Y are produced. This production plan maximizes the producer's total earnings using the available labour and capital. The graph of the feasible region is shown below:
In simple words: The producer should make 3 units of good X and 8 units of good Y. This will use their labor and capital resources efficiently and give them the most money, totaling Rs. 1260.

🎯 Exam Tip: When dealing with resource allocation problems like this, it's essential to correctly identify which variable represents which product and how each product consumes the available resources (labour, capital). Pay close attention to unit conversions if quantities are given in different units.

 

Question 21. A farmer has a supply of chemical fertilizer of type I which contains 10% nitrogen and 6% phosphoric acid and type II fertilizer which contains 5% nitrogen and 10 % phosphoric acid. After testing the soil conditions of a field, it is found that at least 14 kg of nitrogen and 14 kg of phosphoric acid is required for a good crop. The fertilizer type I costs Rs 2.00 per kg and the type II costs Rs 3.00 per kg. How many kilograms of each fertilizer should be used to meet the requirements and the cost be minimum.
Answer: Let \( x \) kg of fertilizer type I and \( y \) kg of fertilizer type II be used. The given information can be summarized in the table below:

The objective is to minimize the total cost \( Z \). So, \( Z = 2x + 3y \). The constraints based on the minimum requirements for nitrogen and phosphoric acid are:
For Nitrogen: \( 0.1x + 0.05y \ge 14 \)
\( \implies \) \( 10x + 5y \ge 1400 \) (multiplying by 100)
\( \implies \) \( 2x + y \ge 280 \) (dividing by 5)
For Phosphoric Acid: \( 0.06x + 0.1y \ge 14 \)
\( \implies \) \( 6x + 10y \ge 1400 \) (multiplying by 100)
\( \implies \) \( 3x + 5y \ge 700 \) (dividing by 2)
Also, the quantity of fertilizer cannot be negative, so \( x \ge 0 \) and \( y \ge 0 \).
We plot the lines \( 2x + y = 280 \) and \( 3x + 5y = 700 \).
For \( 2x + y = 280 \):
If \( x=0 \), \( y=280 \). Point: \( (0, 280) \)
If \( y=0 \), \( x=140 \). Point: \( (140, 0) \)
For \( 3x + 5y = 700 \):
If \( x=0 \), \( y=140 \). Point: \( (0, 140) \)
If \( y=0 \), \( x \approx 233.33 \). Point: \( (700/3, 0) \)
To find the intersection point, solve the system:
\( 2x + y = 280 \Rightarrow y = 280 - 2x \)
Substitute into the second equation: \( 3x + 5(280 - 2x) = 700 \)
\( 3x + 1400 - 10x = 700 \)
\( -7x = -700 \)
\( \implies \) \( x = 100 \)
Then \( y = 280 - 2(100) = 80 \).
The intersection point is \( (100, 80) \).
The feasible region is unbounded and lies above these lines in the first quadrant. The corner points are \( (140, 0) \), \( (100, 80) \), and \( (0, 280) \). Let's evaluate Z at these points:
At \( (140, 0) \): \( Z = 2(140) + 3(0) = 280 \)
At \( (100, 80) \): \( Z = 2(100) + 3(80) = 200 + 240 = 440 \)
At \( (0, 280) \): \( Z = 2(0) + 3(280) = 840 \)
The minimum value of \( Z \) is 280 at \( x = 140 \) and \( y = 0 \). This problem seeks to minimize cost.
In simple words: The farmer should use 140 kg of fertilizer type I and no fertilizer type II. This combination will meet the nutrient needs while keeping the cost as low as possible. Getting the right mix helps plants grow well without wasting money.

🎯 Exam Tip: When minimizing an objective function over an unbounded feasible region, always check if the open half-plane of the minimum Z value has common points with the feasible region. If not, the smallest value obtained at a corner point is indeed the minimum.

 

Question 22. A company produces two types of belts, A and B. Profits on these type are Rs 2 and Rs 1.5 on each belt, respectively. A belt of type A requires twice as much time as a belt of type B. The company C can produce at the most 1000 belts per day, Material for 800 belts per day is available. At the most 400 buckles for belts of type A and 700 for those of type B are available per day. How many belts of each type would the company produce so as to maximize the profit?
Answer: Let \( x \) be the number of belts of type A and \( y \) be the number of belts of type B. The goal is to maximize profit \( Z \).
From the given information, the profit function is \( Z = 2x + 1.5y \).
The constraints are:
1. Production time: A belt of type A takes twice the time of type B. If type B takes 1 unit of time, type A takes 2 units. Total production capacity is 1000 belts per day (meaning total time for 1000 'type B equivalent' units).
\( 2x + y \le 1000 \)
2. Material availability: Enough material for 800 belts per day (total A and B).
\( x + y \le 800 \)
3. Buckles for type A: At most 400 buckles for type A.
\( x \le 400 \)
4. Buckles for type B: At most 700 buckles for type B.
\( y \le 700 \)
5. Non-negativity: \( x \ge 0, y \ge 0 \)
We plot these inequalities on a graph to find the feasible region. The corner points of the feasible region are:
\( O(0,0) \)
\( E_1(400,0) \) (intersection of \( x=400 \) and \( y=0 \))
\( H_1(400,200) \) (intersection of \( x=400 \) and \( 2x+y=1000 \))
\( G_1(200,600) \) (intersection of \( x+y=800 \) and \( 2x+y=1000 \))
\( I_1(100,700) \) (intersection of \( x+y=800 \) and \( y=700 \))
\( F_1(0,700) \) (intersection of \( x=0 \) and \( y=700 \))

We evaluate \( Z = 2x + 1.5y \) at these corner points:
At \( O(0,0) \): \( Z = 2(0) + 1.5(0) = 0 \)
At \( E_1(400,0) \): \( Z = 2(400) + 1.5(0) = 800 \)
At \( H_1(400,200) \): \( Z = 2(400) + 1.5(200) = 800 + 300 = 1100 \)
At \( G_1(200,600) \): \( Z = 2(200) + 1.5(600) = 400 + 900 = 1300 \)
At \( I_1(100,700) \): \( Z = 2(100) + 1.5(700) = 200 + 1050 = 1250 \)
At \( F_1(0,700) \): \( Z = 2(0) + 1.5(700) = 1050 \)
The maximum profit \( Z \) is Rs 1300, which occurs at \( x = 200 \) and \( y = 600 \).
In simple words: To get the most profit, the company should make 200 belts of type A and 600 belts of type B. This plan makes sure they use their machines and materials wisely, leading to the biggest earnings.

🎯 Exam Tip: Always clearly label the axes and constraint lines on your graph. Identify all corner points of the feasible region and evaluate the objective function at each to find the maximum or minimum value.

 

Question 23. An oil company requires 12,000, 20,000 and 15,000 barrels of high-grade, medium grade and low grade oil, respectively. Refinery A produces 100,300 and 200 barrels per day of high-grade, medium-grade and low-grade oil, respectively, while refinery B produces 200, 400 and 100 barrels per day of high-grade, medium-grade and low-grade oil, respectively. If refinery A costs Rs 400 per day and refinery B costs Rs 300 per day to operate, how many days should each be run to minimize costs while satisfying requirements.
Answer: Let \( x \) be the number of days Refinery A is run and \( y \) be the number of days Refinery B is run. The goal is to minimize the total cost \( Z \).
The given data can be arranged in a table:

The objective function to minimize cost is \( Z = 400x + 300y \).
The constraints are:
1. High-grade oil: \( 100x + 200y \ge 12000 \)
\( \implies \) \( x + 2y \ge 120 \) (dividing by 100)
2. Medium-grade oil: \( 300x + 400y \ge 20000 \)
\( \implies \) \( 3x + 4y \ge 200 \) (dividing by 100)
3. Low-grade oil: \( 200x + 100y \ge 15000 \)
\( \implies \) \( 2x + y \ge 150 \) (dividing by 100)
4. Non-negativity: \( x \ge 0, y \ge 0 \)
We plot the lines and find the feasible region. This region is unbounded. The corner points of the feasible region are:
\( F(0,150) \) (from \( 2x+y=150 \))
\( P(60,30) \) (intersection of \( x+2y=120 \) and \( 2x+y=150 \))
\( A(120,0) \) (from \( x+2y=120 \))

We evaluate \( Z = 400x + 300y \) at these corner points:
At \( A(120, 0) \): \( Z = 400(120) + 300(0) = 48000 \)
At \( P(60, 30) \): \( Z = 400(60) + 300(30) = 24000 + 9000 = 33000 \)
At \( F(0, 150) \): \( Z = 400(0) + 300(150) = 45000 \)
The minimum cost is Rs 33000, which occurs at \( x = 60 \) and \( y = 30 \).
In simple words: To meet all the oil requirements at the lowest possible cost, Refinery A should run for 60 days, and Refinery B should run for 30 days. This schedule ensures enough oil is produced without overspending.

🎯 Exam Tip: For minimization problems with an unbounded feasible region, always test whether the lowest Z-value found at a corner point is truly the minimum by drawing the objective function line for that Z-value and checking the open half-plane for intersection with the feasible region.

 

Question 24. A company produces soft drinks that has a contract which requires that a minimum of 80 units of the chemical A and 60 units of the chemical B to go into each bottle of the drink. The chemicals are available in a prepared mix from two different suppliers. Supplier S has a mix of 4 units of A and 2 units of B that costs Rs 10, the supplier T has a mix of 1 unit of A and 1 unit of B that costs Rs 4. How many mixes from S and T should the company purchase to honour contract requirement and yet minimize cost?
Answer: Let \( x \) be the number of mixes purchased from Supplier S and \( y \) be the number of mixes purchased from Supplier T. The goal is to minimize the total cost \( Z \).
The given data can be arranged in a table:

The objective function to minimize cost is \( Z = 10x + 4y \).
The constraints are:
1. Chemical A: \( 4x + y \ge 80 \)
2. Chemical B: \( 2x + y \ge 60 \)
3. Non-negativity: \( x \ge 0, y \ge 0 \)
We plot the lines and find the feasible region. This region is unbounded. The corner points of the feasible region are:
\( B(0,80) \) (from \( 4x+y=80 \))
\( P(10,40) \) (intersection of \( 4x+y=80 \) and \( 2x+y=60 \))
\( C(30,0) \) (from \( 2x+y=60 \))

We evaluate \( Z = 10x + 4y \) at these corner points:
At \( B(0,80) \): \( Z = 10(0) + 4(80) = 320 \)
At \( P(10,40) \): \( Z = 10(10) + 4(40) = 100 + 160 = 260 \)
At \( C(30,0) \): \( Z = 10(30) + 4(0) = 300 \)
The minimum cost is Rs 260, which occurs at \( x = 10 \) and \( y = 40 \).
In simple words: To meet the chemical requirements at the lowest cost, the company should buy 10 mixes from Supplier S and 40 mixes from Supplier T. This way, they get all the necessary ingredients without overspending.

🎯 Exam Tip: Remember to always verify the minimum cost for unbounded regions by checking if the open half-plane for that Z-value intersects the feasible region. If it does not, the point is indeed the minimum.

Examples

 

Question 1. A manufacturer produces two types of steel trunks. He has two machines, A and B, The first type trunk requires 3 hours on machine A and 3 hours on machine B. The second type trunk requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs 30 per trunk on the first type of trunk and Rs 25 per trunk on the second type. Formulate a linear programming problem to find out how many trunks of each type he must make each day to maximize his profit.
Answer: Let \( x \) be the number of trunks of type I and \( y \) be the number of trunks of type II. The goal is to maximize the total profit \( Z \).
The objective function to maximize profit is \( Z = 30x + 25y \).
The constraints are:
1. Machine A time: Type I needs 3 hours, Type II needs 3 hours. Max 18 hours.
\( 3x + 3y \le 18 \)
\( \implies \) \( x + y \le 6 \) (dividing by 3)
2. Machine B time: Type I needs 3 hours, Type II needs 2 hours. Max 15 hours.
\( 3x + 2y \le 15 \)
3. Non-negativity: \( x \ge 0, y \ge 0 \)
We plot the lines for these inequalities. The feasible region is bounded by these lines and the axes. The corner points of the feasible region are:
\( O(0,0) \)
\( C_1(5,0) \) (from \( 3x+2y=15 \))
\( E_1(3,3) \) (intersection of \( x+y=6 \) and \( 3x+2y=15 \))
\( B_1(0,6) \) (from \( x+y=6 \))

We evaluate \( Z = 30x + 25y \) at these corner points:
At \( O(0,0) \): \( Z = 30(0) + 25(0) = 0 \)
At \( C_1(5,0) \): \( Z = 30(5) + 25(0) = 150 \)
At \( E_1(3,3) \): \( Z = 30(3) + 25(3) = 90 + 75 = 165 \)
At \( B_1(0,6) \): \( Z = 30(0) + 25(6) = 150 \)
The maximum profit is Rs 165, which occurs at \( x = 3 \) and \( y = 3 \).
In simple words: To get the most profit, the manufacturer should produce 3 trunks of type I and 3 trunks of type II each day. This balance ensures both machines are used efficiently and profit is maximized.

🎯 Exam Tip: When dealing with machine time or resource constraints, always convert all time units to a common unit (e.g., minutes or hours) to avoid errors in formulating the inequalities.

 

Question 2. A new cereal, formed of a mixture of bran and rice, contains at least 88 grams of protein and at east 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligram of iron per kilogram, and that rice contains 100 grams of protein and 30 milligram of iron per kilogram, find the minimum cost of producing a kilogram of this new cereal if bran costs Rs 28 per kilogram and rice costs Rs 25 per kilogram.
Answer: Let \( x \) kg of bran and \( y \) kg of rice be used in the cereal. The goal is to minimize the total cost \( Z \).
The objective function to minimize cost is \( Z = 28x + 25y \).
The constraints are:
1. Protein requirement: At least 88 grams.
\( 80x + 100y \ge 88 \)
\( \implies \) \( 20x + 25y \ge 22 \) (dividing by 4)
2. Iron requirement: At least 36 milligrams.
\( 40x + 30y \ge 36 \)
\( \implies \) \( 20x + 15y \ge 18 \) (dividing by 2)
3. Non-negativity: \( x \ge 0, y \ge 0 \)
We plot the lines for these inequalities. The feasible region is unbounded and lies above these lines in the first quadrant. The corner points of the feasible region are:
\( A(1.1,0) \) (from \( 20x+25y=22 \))
\( E(0.6,0.4) \) (intersection of \( 20x+25y=22 \) and \( 20x+15y=18 \))
\( D(0,1.2) \) (from \( 20x+15y=18 \))

We evaluate \( Z = 28x + 25y \) at these corner points:
At \( A(1.1, 0) \): \( Z = 28(1.1) + 25(0) = 30.80 \)
At \( E(0.6, 0.4) \): \( Z = 28(0.6) + 25(0.4) = 16.8 + 10 = 26.80 \)
At \( D(0, 1.2) \): \( Z = 28(0) + 25(1.2) = 30 \)
The minimum cost is Rs 26.80, which occurs at \( x = 0.6 \) and \( y = 0.4 \).
In simple words: To make the cereal at the lowest cost, you need to mix 0.6 kg of bran and 0.4 kg of rice. This blend provides the right amount of protein and iron without spending too much money.

🎯 Exam Tip: When simplifying constraint equations for graphing, ensure you apply the division consistently across all terms, including the constant, to avoid errors in plotting the lines.

 

Question 2. A new cereal, formed of a mixture of bran and rice, contains at least 88 grams of protein and at east 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligram of iron per kilogram, and that rice contains 100 grams of protein and 30 milligram of iron per kilogram, find the minimum cost of producing a kilogram of this new cereal if bran costs 28 per kilogram and rice costs 25 per kilogram.
Answer: Let \(x\) kg be the quantity of bran and \(y\) kg be the quantity of rice in the cereal mixture. Our goal is to minimize the total cost. The cost for \(x\) kg of bran is Rs 28\(x\) and for \(y\) kg of rice is Rs 25\(y\). So, the total cost \(Z = 28x + 25y\).
We have the following constraints based on the minimum requirements for protein and iron:
For protein: \(80x + 100y \ge 88\)
\( \implies \) \(20x + 25y \ge 22\) (This simplifies by dividing by 4).
For iron: \(40x + 30y \ge 36\).
Since quantities cannot be negative, we also have \(x \ge 0\) and \(y \ge 0\). Linear programming helps find the best way to use limited resources to achieve a goal, like minimizing cost in this case.
We convert the inequalities into equations to find the corner points of the feasible region:
1. \(20x + 25y = 22\). This line intersects the axes at \(A(\frac{11}{10}, 0)\) or \((1.1, 0)\), and \(B(0, \frac{22}{25})\) or \((0, 0.88)\).
2. \(40x + 30y = 36\). This line intersects the axes at \(C(\frac{9}{10}, 0)\) or \((0.9, 0)\), and \(D(0, \frac{6}{5})\) or \((0, 1.2)\).
The intersection of \(20x + 25y = 22\) and \(40x + 30y = 36\) (which is \(20x + 15y = 18\)) is found by solving the system:
\(20x + 25y = 22\) (i)
\(20x + 15y = 18\) (ii)
Subtracting (ii) from (i): \(10y = 4 \implies y = \frac{4}{10} = 0.4\)
Substitute \(y = 0.4\) into (ii): \(20x + 15(0.4) = 18 \implies 20x + 6 = 18 \implies 20x = 12 \implies x = \frac{12}{20} = 0.6\).
So, the intersection point is \(E(0.6, 0.4)\).

Now, we evaluate the objective function \(Z = 28x + 25y\) at the corner points of the feasible region (A, E, D):
\(A(\frac{11}{10}, 0)\) or \((1.1, 0)\): \(Z = 28(1.1) + 25(0) = 30.80\)
\(E(\frac{3}{5}, \frac{2}{5})\) or \((0.6, 0.4)\): \(Z = 28(0.6) + 25(0.4) = 16.8 + 10 = 26.80\)
\(D(0, \frac{6}{5})\) or \((0, 1.2)\): \(Z = 28(0) + 25(1.2) = 30\)
The smallest value of \(Z\) is 26.80 at \(x = 0.6\) and \(y = 0.4\). Since the feasible region is unbounded, we check if the open half-plane \(28x + 25y < 26.80\) has any common points with the feasible region. It does not. Therefore, 26.80 is indeed the minimum cost.
Thus, to minimize the cost, the mixture should contain 0.6 kg of bran and 0.4 kg of rice, resulting in a minimum cost of Rs 26.80.
In simple words: To make the cereal at the lowest cost, you should use 0.6 kilograms of bran and 0.4 kilograms of rice. This will cost Rs 26.80, which is the cheapest way to meet all the protein and iron needs.

🎯 Exam Tip: For problems involving minimum requirements, constraints will often use 'greater than or equal to' (\( \ge \)), leading to an unbounded feasible region. Always check the objective function value on the corners, and if the region is unbounded, graph a line for the minimum value (Z = constant) to confirm no part of the feasible region lies below it.

 

Question 3. A company produces two types of items, P and Q. Manufacturing of both items requires the metals gold and copper. Each unit of item P requires 3 grams of gold and 1 gram of copper while that of item Q requires 1 gram of gold and 2 grams of copper. The company has 9 grams of gold and 8 grams of copper in store. If each unit of item P makes a profit of Rs 50 and each unit of item Q makes a profit of Rs 60, determine the number of units of each item that the company should produce to maximize profit. What is the maximum profit ?
Answer: Let \(x\) be the number of units of item P and \(y\) be the number of units of item Q produced. We want to maximize the total profit.
The profit from each unit of item P is Rs 50, so for \(x\) units, the profit is Rs \(50x\).
The profit from each unit of item Q is Rs 60, so for \(y\) units, the profit is Rs \(60y\).
The total profit \(Z = 50x + 60y\). This is the objective function we need to maximize.
Now, let's look at the constraints based on the available metals:
For gold: Each unit of P needs 3g, and each unit of Q needs 1g. Total gold available is 9g.
So, \(3x + 1y \le 9\).
For copper: Each unit of P needs 1g, and each unit of Q needs 2g. Total copper available is 8g.
So, \(1x + 2y \le 8\).
Since we cannot produce a negative number of items, we have \(x \ge 0\) and \(y \ge 0\). This also means our feasible region will be in the first quadrant. To find the maximum profit, we will check the profit at the corner points of the feasible region.

We convert the inequalities to equations to find the intersection points:
1. \(3x + y = 9\). This line intersects the axes at \((3, 0)\) and \((0, 9)\).
2. \(x + 2y = 8\). This line intersects the axes at \((8, 0)\) and \((0, 4)\).

To find the intersection of these two lines, we solve the system of equations:
\(3x + y = 9 \implies y = 9 - 3x\) (i)
\(x + 2y = 8\) (ii)
Substitute \(y\) from (i) into (ii):
\(x + 2(9 - 3x) = 8\)
\(x + 18 - 6x = 8\)
\(-5x = 8 - 18\)
\(-5x = -10\)
\(x = 2\)
Substitute \(x = 2\) back into (i): \(y = 9 - 3(2) = 9 - 6 = 3\).
So, the intersection point is \(P(2, 3)\).

The corner points of the feasible region are O(0,0), A(3,0), P(2,3), and D(0,4). We evaluate the objective function \(Z = 50x + 60y\) at these points:
O(0,0): \(Z = 50(0) + 60(0) = 0\)
A(3,0): \(Z = 50(3) + 60(0) = 150\)
P(2,3): \(Z = 50(2) + 60(3) = 100 + 180 = 280\)
D(0,4): \(Z = 50(0) + 60(4) = 240\)
The maximum profit is Rs 280, which occurs when \(x = 2\) and \(y = 3\). This means the company should produce 2 units of item P and 3 units of item Q to get the highest profit. This optimization helps businesses make the most out of their resources.
In simple words: To make the most money, the company should make 2 units of product P and 3 units of product Q. This will give them the biggest profit, which is Rs 280.

🎯 Exam Tip: In optimization problems, the feasible region is often a polygon. The maximum or minimum value of the objective function will always occur at one of the vertices (corner points) of this polygon.

 

Question 4. A manufacturer manufactures two types of tea-cups, A and B. Three machines are needed for manufacturing the tea cups. The time in minutes required for manufacturing each cup on the machines is given below:
(i) Type of cup A on Machine I: 12 minutes, Machine II: 18 minutes, Machine III: 6 minutes
(ii) Type of cup B on Machine I: 6 minutes, Machine II: 0 minutes, Machine III: 9 minutes
Each machine is available for a maximum of 6 hours per day. If the profit on each cup of type A is Rs 1.50 and that on each cup of type B is Rs 1.00, find the number of cups of each type that should be manufactured in a day to get maximum profit.
Answer: Let \(x\) be the number of tea-cups of type A and \(y\) be the number of tea-cups of type B manufactured per day. We want to maximize the profit.
The profit for each cup of type A is Rs 1.50, so for \(x\) cups, it's \(1.50x\).
The profit for each cup of type B is Rs 1.00, so for \(y\) cups, it's \(1.00y\).
The total profit (objective function) is \(Z = 1.50x + 1.00y\).

The machines are available for a maximum of 6 hours each day, which is \(6 \times 60 = 360\) minutes.
Constraints based on machine time:
Machine I: 12 minutes for A, 6 minutes for B. Max 360 minutes.
So, \(12x + 6y \le 360\), which simplifies to \(2x + y \le 60\) (dividing by 6).
Machine II: 18 minutes for A, 0 minutes for B. Max 360 minutes.
So, \(18x + 0y \le 360\), which simplifies to \(x \le 20\).
Machine III: 6 minutes for A, 9 minutes for B. Max 360 minutes.
So, \(6x + 9y \le 360\), which simplifies to \(2x + 3y \le 120\) (dividing by 3).
Also, the number of cups cannot be negative, so \(x \ge 0\) and \(y \ge 0\). This means the feasible region is in the first quadrant.

To find the corner points of the feasible region, we examine the boundary lines:
1. \(2x + y = 60\). Intersects axes at \((30, 0)\) and \((0, 60)\).
2. \(x = 20\). A vertical line through \(x = 20\).
3. \(2x + 3y = 120\). Intersects axes at \((60, 0)\) and \((0, 40)\).

Now, we find the intersection points of these lines:
- Intersection of \(2x + y = 60\) and \(x = 20\):
Substitute \(x = 20\) into \(2x + y = 60\): \(2(20) + y = 60 \implies 40 + y = 60 \implies y = 20\). Point H\(_{1}\)(20, 20).
- Intersection of \(2x + y = 60\) and \(2x + 3y = 120\):
Subtract \(2x + y = 60\) from \(2x + 3y = 120\): \(2y = 60 \implies y = 30\).
Substitute \(y = 30\) into \(2x + y = 60\): \(2x + 30 = 60 \implies 2x = 30 \implies x = 15\). Point G\(_{1}\)(15, 30).

The corner points of the feasible region O(0,0), C\(_{1}\)(20,0), H\(_{1}\)(20,20), G\(_{1}\)(15,30), and F\(_{1}\)(0,40). We evaluate \(Z = 1.50x + 1.00y\) at these points:
O(0,0): \(Z = 1.50(0) + 1.00(0) = 0\)
C\(_{1}\)(20,0): \(Z = 1.50(20) + 1.00(0) = 30\)
H\(_{1}\)(20,20): \(Z = 1.50(20) + 1.00(20) = 30 + 20 = 50\)
G\(_{1}\)(15,30): \(Z = 1.50(15) + 1.00(30) = 22.5 + 30 = 52.5\)
F\(_{1}\)(0,40): \(Z = 1.50(0) + 1.00(40) = 40\)
The maximum profit is Rs 52.50, achieved when \(x = 15\) and \(y = 30\). This means the manufacturer should produce 15 cups of type A and 30 cups of type B to maximize profit. This helps in efficient resource allocation.
In simple words: To get the most profit, the factory should make 15 cups of type A and 30 cups of type B. This will give them a total profit of Rs 52.50.

🎯 Exam Tip: When dealing with multiple constraints (like different machines with limited hours), draw each constraint line and shade the allowed region. The feasible region for a maximization problem will be the area where all shaded regions overlap, and its vertices give the potential optimal solutions.

 

Question 5. Two tailors P and Q earn Rs. 150 and Rs. 200 respectively. P can stitch 6 shirts and 4 trousers a day, while Q can stitch 10 shirts and 4 trousers per day. How many days should each work to produce at least 60 shirts and 32 trousers at minimum labour cost?
Answer: Let \(x\) be the number of days tailor P works and \(y\) be the number of days tailor Q works. We want to minimize the total labor cost.
Tailor P earns Rs 150 per day, so for \(x\) days, the cost is \(150x\).
Tailor Q earns Rs 200 per day, so for \(y\) days, the cost is \(200y\).
The total labor cost (objective function) is \(Z = 150x + 200y\).

Now, let's establish the constraints based on the required production:
Shirts: P stitches 6 shirts/day, Q stitches 10 shirts/day. At least 60 shirts are needed.
So, \(6x + 10y \ge 60\), which simplifies to \(3x + 5y \ge 30\) (dividing by 2).
Trousers: P stitches 4 trousers/day, Q stitches 4 trousers/day. At least 32 trousers are needed.
So, \(4x + 4y \ge 32\), which simplifies to \(x + y \ge 8\) (dividing by 4).
Since the number of days cannot be negative, we have \(x \ge 0\) and \(y \ge 0\). This means the feasible region is in the first quadrant. Finding the optimal solution here helps a business keep costs down.

To find the corner points of the feasible region, we look at the boundary lines:
1. \(3x + 5y = 30\). Intersects axes at \((10, 0)\) and \((0, 6)\).
2. \(x + y = 8\). Intersects axes at \((8, 0)\) and \((0, 8)\).

Now, we find the intersection point of these two lines:
From \(x + y = 8 \implies y = 8 - x\).
Substitute into \(3x + 5y = 30\):
\(3x + 5(8 - x) = 30\)
\(3x + 40 - 5x = 30\)
\(-2x = 30 - 40\)
\(-2x = -10\)
\(x = 5\)
Substitute \(x = 5\) back into \(y = 8 - x\): \(y = 8 - 5 = 3\).
So, the intersection point is \(E_{1}(5, 3)\).

The corner points of the feasible region are A\(_{1}\)(10,0), E\(_{1}\)(5,3), and D\(_{1}\)(0,8). We evaluate the objective function \(Z = 150x + 200y\) at these points:
A\(_{1}\)(10,0): \(Z = 150(10) + 200(0) = 1500\)
E\(_{1}\)(5,3): \(Z = 150(5) + 200(3) = 750 + 600 = 1350\)
D\(_{1}\)(0,8): \(Z = 150(0) + 200(8) = 1600\)
The smallest value of \(Z\) is 1350 at \(x = 5\) and \(y = 3\). Since the feasible region is unbounded (as constraints are \( \ge \)), we must verify this is the minimum. We draw the line \(15x + 20y = 135\). The open half-plane \(15x + 20y < 135\) has no points in common with the feasible region, confirming 1350 is the minimum.
Thus, tailor P should work for 5 days and tailor Q should work for 3 days to produce the required items at a minimum cost of Rs 1350.
In simple words: To pay the least amount, tailor P should work for 5 days and tailor Q should work for 3 days. The total cost will be Rs 1350.

🎯 Exam Tip: When dealing with minimization problems with 'at least' requirements, the feasible region is usually unbounded. Always remember to check the region formed by the objective function's minimum value line to ensure it doesn't overlap with the feasible region, confirming the true minimum.

 

Question 6. A mill owner buys two types of machines A and B for his mill. Machine A occupies 1000 sqm of area and requires 12 men to operate it; while machine B occupies 1200 sqm of area and requires 8 men to operate it. The owner has 7600 sqm of area available and 72 men to operate the machines. If machine A produces 50 units and machine B produces 40 units daily, how many machines of each type should he buy to maximise the daily output? Use Linear Programming to find the solution.
Answer: Let \(x\) be the number of machines of type A and \(y\) be the number of machines of type B that the mill owner buys. The goal is to maximize the daily output.
Machine A produces 50 units daily, so \(x\) machines produce \(50x\) units.
Machine B produces 40 units daily, so \(y\) machines produce \(40y\) units.
The total daily output (objective function) is \(Z = 50x + 40y\).

Now, let's establish the constraints based on available area and manpower:
Area constraint: Machine A needs 1000 sqm, Machine B needs 1200 sqm. Total area available is 7600 sqm.
So, \(1000x + 1200y \le 7600\), which simplifies to \(5x + 6y \le 38\) (dividing by 200).
Manpower constraint: Machine A needs 12 men, Machine B needs 8 men. Total men available is 72.
So, \(12x + 8y \le 72\), which simplifies to \(3x + 2y \le 18\) (dividing by 4).
Since the number of machines cannot be negative, we have \(x \ge 0\) and \(y \ge 0\). This means the feasible region is in the first quadrant. Businesses use this kind of planning to make sure they get the most production from their limited resources.

To find the corner points of the feasible region, we examine the boundary lines:
1. \(5x + 6y = 38\). Intersects axes at \((\frac{38}{5}, 0)\) or \((7.6, 0)\) and \((0, \frac{38}{6})\) or \((0, \frac{19}{3} \approx 6.33)\).
2. \(3x + 2y = 18\). Intersects axes at \((6, 0)\) and \((0, 9)\).

Now, we find the intersection point of these two lines:
Multiply \(3x + 2y = 18\) by 3 to match the \(y\) coefficient of the first equation:
\(9x + 6y = 54\) (ii')
Subtract \(5x + 6y = 38\) (i) from \(9x + 6y = 54\) (ii'):
\((9x + 6y) - (5x + 6y) = 54 - 38\)
\(4x = 16 \implies x = 4\)
Substitute \(x = 4\) into \(3x + 2y = 18\):
\(3(4) + 2y = 18 \implies 12 + 2y = 18 \implies 2y = 6 \implies y = 3\).
So, the intersection point is P(4, 3).

The corner points of the feasible region are O(0,0), C(6,0), P(4,3), and B\((0, \frac{19}{3})\). We evaluate the objective function \(Z = 50x + 40y\) at these points:
O(0,0): \(Z = 50(0) + 40(0) = 0\)
C(6,0): \(Z = 50(6) + 40(0) = 300\)
P(4,3): \(Z = 50(4) + 40(3) = 200 + 120 = 320\)
B\((0, \frac{19}{3})\): \(Z = 50(0) + 40(\frac{19}{3}) = \frac{760}{3} \approx 253.33\)
The maximum output is 320 units, achieved when \(x = 4\) and \(y = 3\). This means the mill owner should buy 4 machines of type A and 3 machines of type B to get the maximum daily output.
In simple words: To get the most products from the mill, the owner should buy 4 machines of type A and 3 machines of type B. This way, they can make 320 units every day.

🎯 Exam Tip: When setting up constraints for linear programming, ensure all units are consistent (e.g., all time in minutes or all in hours). This prevents errors in calculations and correctly defines the feasible region.

 

Question 7. A company manufactures two types of toys A and B. A toy of type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is Rs 50 each on type A and Rs 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit? Use linear programming to find the solution.
Answer: Let \(x\) be the number of toys of type A and \(y\) be the number of toys of type B manufactured per day. We want to maximize the profit.
The profit for each toy of type A is Rs 50, so for \(x\) toys, it's \(50x\).
The profit for each toy of type B is Rs 60, so for \(y\) toys, it's \(60y\).
The total profit (objective function) is \(Z = 50x + 60y\).

Now, let's establish the constraints based on available machine time:
Available time for cutting: 3 hours = \(3 \times 60 = 180\) minutes.
Cutting constraint: Toy A needs 5 min, Toy B needs 8 min.
So, \(5x + 8y \le 180\).
Available time for assembling: 4 hours = \(4 \times 60 = 240\) minutes.
Assembling constraint: Toy A needs 10 min, Toy B needs 8 min.
So, \(10x + 8y \le 240\), which simplifies to \(5x + 4y \le 120\) (dividing by 2).
Also, the number of toys cannot be negative, so \(x \ge 0\) and \(y \ge 0\). This means the feasible region is in the first quadrant. Careful planning helps in optimizing production when there are different steps involved.

To find the corner points of the feasible region, we examine the boundary lines:
1. \(5x + 8y = 180\). Intersects axes at \((\frac{180}{5}, 0)\) or \((36, 0)\) and \((0, \frac{180}{8})\) or \((0, 22.5)\).
2. \(5x + 4y = 120\). Intersects axes at \((\frac{120}{5}, 0)\) or \((24, 0)\) and \((0, \frac{120}{4})\) or \((0, 30)\).

Now, we find the intersection point of these two lines:
Subtract \(5x + 4y = 120\) (ii) from \(5x + 8y = 180\) (i):
\((5x + 8y) - (5x + 4y) = 180 - 120\)
\(4y = 60 \implies y = 15\)
Substitute \(y = 15\) into \(5x + 4y = 120\):
\(5x + 4(15) = 120 \implies 5x + 60 = 120 \implies 5x = 60 \implies x = 12\).
So, the intersection point is E\(_{1}\)(12, 15).

The corner points of the feasible region are O(0,0), C\(_{1}\)(24,0), E\(_{1}\)(12,15), and B\(_{1}\)(0, 22.5). We evaluate the objective function \(Z = 50x + 60y\) at these points:
O(0,0): \(Z = 50(0) + 60(0) = 0\)
C\(_{1}\)(24,0): \(Z = 50(24) + 60(0) = 1200\)
E\(_{1}\)(12,15): \(Z = 50(12) + 60(15) = 600 + 900 = 1500\)
B\(_{1}\)(0, 22.5): \(Z = 50(0) + 60(22.5) = 1350\)
The maximum profit is Rs 1500, achieved when \(x = 12\) and \(y = 15\). This means the company should manufacture 12 toys of type A and 15 toys of type B to earn the maximum profit.
In simple words: To get the most profit, the company should make 12 toys of type A and 15 toys of type B. This will bring them the biggest profit of Rs 1500.

🎯 Exam Tip: When visualizing the feasible region, ensure you correctly identify the 'less than or equal to' (\( \le \)) or 'greater than or equal to' (\( \ge \)) side of each line. For 'less than or equal to', the region is typically towards the origin if the origin satisfies the inequality.

 

Question 8. A dietician wishes to mix two kinds of food X and Tin such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
(i) Food X: Vitamin A (1 unit), Vitamin B (2 units), Vitamin C (3 units)
(ii) Food Y: Vitamin A (2 units), Vitamin B (2 units), Vitamin C (1 unit)
One kg of food X costs Rs. 24 and one kg of food F cost Rs. 36. Using Linear Programming, find the least cost of the total mixture which will contain the required vitamins.
Answer: Let \(x\) kg be the amount of food X and \(y\) kg be the amount of food Y in the mixture. We want to minimize the total cost.
The cost of 1 kg of food X is Rs 24, so for \(x\) kg, it's \(24x\).
The cost of 1 kg of food Y is Rs 36, so for \(y\) kg, it's \(36y\).
The total cost (objective function) is \(Z = 24x + 36y\).

Now, let's establish the constraints based on the minimum vitamin requirements:
Vitamin A: Food X provides 1 unit/kg, Food Y provides 2 units/kg. At least 10 units are needed.
So, \(x + 2y \ge 10\).
Vitamin B: Food X provides 2 units/kg, Food Y provides 2 units/kg. At least 12 units are needed.
So, \(2x + 2y \ge 12\), which simplifies to \(x + y \ge 6\) (dividing by 2).
Vitamin C: Food X provides 3 units/kg, Food Y provides 1 unit/kg. At least 8 units are needed.
So, \(3x + y \ge 8\).
Also, the amount of food cannot be negative, so \(x \ge 0\) and \(y \ge 0\). This means the feasible region is in the first quadrant. Dieticians use this to ensure nutritional requirements are met at the lowest cost.

To find the corner points of the feasible region, we examine the boundary lines:
1. \(x + 2y = 10\). Intersects axes at \((10, 0)\) and \((0, 5)\).
2. \(x + y = 6\). Intersects axes at \((6, 0)\) and \((0, 6)\).
3. \(3x + y = 8\). Intersects axes at \((\frac{8}{3}, 0)\) or \((2.67, 0)\) and \((0, 8)\).

Now, we find the intersection points of these lines:
- Intersection of \(x + 2y = 10\) and \(x + y = 6\):
Subtract \(x + y = 6\) from \(x + 2y = 10\): \(y = 4\).
Substitute \(y = 4\) into \(x + y = 6\): \(x + 4 = 6 \implies x = 2\). Point H(2, 4).
- Intersection of \(x + y = 6\) and \(3x + y = 8\):
Subtract \(x + y = 6\) from \(3x + y = 8\): \(2x = 2 \implies x = 1\).
Substitute \(x = 1\) into \(x + y = 6\): \(1 + y = 6 \implies y = 5\). Point G(1, 5).

The corner points of the feasible region are A(10,0), H(2,4), G(1,5), and F(0,8). We evaluate the objective function \(Z = 24x + 36y\) at these points:
A(10,0): \(Z = 24(10) + 36(0) = 240\)
H(2,4): \(Z = 24(2) + 36(4) = 48 + 144 = 192\)
G(1,5): \(Z = 24(1) + 36(5) = 24 + 180 = 204\)
F(0,8): \(Z = 24(0) + 36(8) = 288\)
The smallest value of \(Z\) is 192, which occurs when \(x = 2\) and \(y = 4\). The feasible region is unbounded, but checking the open half-plane \(24x + 36y < 192\) reveals no common points with the feasible region, so 192 is the true minimum.
Thus, 2 kg of food X and 4 kg of food Y will be required to minimize the cost of the diet, with a minimum cost of Rs 192.
In simple words: To make the mixture with the lowest cost, you need to use 2 kg of food X and 4 kg of food Y. This will cost Rs 192, and it meets all the vitamin needs.

🎯 Exam Tip: For problems with 'at least' conditions and an unbounded feasible region, the minimum value is found at a corner point. Always verify this by checking if the region formed by the objective function's value line \(Z = k\) and its open half-plane \(Z < k\) has no common points with the feasible region.

 

Question 8. A dietician wishes to mix two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

One kg of food X costs Rs. 24 and one kg of food Y cost Rs. 36. Using Linear Programming, find the least cost of the total mixture which will contain the required vitamins.
Answer: Let \(x\) kg of food X and \(y\) kg of food Y be used. The total cost, which we want to minimize, is \( Z = 24x + 36y \). We have constraints based on the minimum vitamin requirements:
For Vitamin A: \( x + 2y \ge 10 \)
For Vitamin B: \( 2x + 2y \ge 12 \implies x + y \ge 6 \)
For Vitamin C: \( 3x + y \ge 8 \)
Also, the quantities of food cannot be negative, so \( x \ge 0, y \ge 0 \).
We find the corner points of the feasible region formed by these inequalities: A(10, 0), H(2, 4), G(1, 5), and F(0, 8). These points represent where the boundary lines cross.
Now, we check the total cost (Z) at each corner point:
At A(10, 0): \( Z = 24(10) + 36(0) = 240 \)
At H(2, 4): \( Z = 24(2) + 36(4) = 48 + 144 = 192 \)
At G(1, 5): \( Z = 24(1) + 36(5) = 24 + 180 = 204 \)
At F(0, 8): \( Z = 24(0) + 36(8) = 288 \)
The minimum cost is Rs. 192, which occurs when 2 kg of food X and 4 kg of food Y are mixed. This blend meets all the necessary vitamin amounts for the lowest price.In simple words: To get enough vitamins with the lowest cost, the dietician should use 2 kg of food X and 4 kg of food Y, which will cost Rs. 192. This specific mix ensures all health requirements are met economically.

🎯 Exam Tip: For unbounded feasible regions, always plot the objective function line for the minimum value and verify it has no points in common with the feasible region to confirm the minimum exists.

 

Question 9. A company manufactures, two types of products A and B. Each unit of A requires 3 grams of nickel and 1 gram of chromium, while each unit of B requires 1 gram of nickel and 2 grams of chromium. The firm can produce 9 grams of nickel and 8 grams of chromium. The profit is Rs. 40 on each unit of product of type A and Rs. 50 on each unit of product of type B. How many units of each type should the company manufacture so as to earn maximum profit? Use linear programming to find the solution.
Answer: Let \(x\) be the number of units of product A and \(y\) be the number of units of product B. The company wants to maximize profit, so the objective function is \( Z = 40x + 50y \).
The constraints based on available materials are:
For nickel: \( 3x + y \le 9 \)
For chromium: \( x + 2y \le 8 \)
Since the number of products cannot be negative, we also have \( x \ge 0, y \ge 0 \).
We identify the corner points of the feasible region defined by these constraints. These points include O(0,0), A(3,0), P(2,3), and D(0,4). The point P(2,3) is where the lines \( 3x + y = 9 \) and \( x + 2y = 8 \) intersect.
Now, we calculate the profit (Z) at each corner point:
At O(0,0): \( Z = 40(0) + 50(0) = 0 \)
At A(3,0): \( Z = 40(3) + 50(0) = 120 \)
At P(2,3): \( Z = 40(2) + 50(3) = 80 + 150 = 230 \)
At D(0,4): \( Z = 40(0) + 50(4) = 200 \)
The maximum profit is Rs. 230, which is achieved when the company manufactures 2 units of product A and 3 units of product B. This strategy allows the company to make the most money given its resource limits.
In simple words: To make the most profit, the company should produce 2 items of type A and 3 items of type B. This will give them a total profit of Rs. 230, which is the highest possible.

🎯 Exam Tip: To ensure maximum profit in Linear Programming Problems, always evaluate the objective function at all corner points of the feasible region, as the optimal solution will lie at one of these vertices.