Samacheer Kalvi Class 11 Business Maths Solutions Chapter 10 Operations Research Exercise 10.1

Question 1. A company produces two types of pens A and B. Pen A is of superior quality and pen B is of lower quality. Profits on pens A and B are Rs 5 and Rs 3 per pen respectively. Raw materials required for each pen A is twice as that of pen B. The supply of raw material is sufficient only for 1000 pens per day. Pen A requires a special clip and only 400 such clips are available per day. For pen B, only 700 clips are available per day. Formulate this problem as a linear programming problem.
Answer:
(i) Variables:
Let \( x_1 \) be the number of pens of type A.
Let \( x_2 \) be the number of pens of type B.

(ii) Objective function:
The profit on \( x_1 \) pens of type A is \( 5x_1 \).
The profit on \( x_2 \) pens of type B is \( 3x_2 \).
Total profit \( Z = 5x_1 + 3x_2 \).
The goal is to maximize this total profit, so the objective function is to maximize \( Z = 5x_1 + 3x_2 \). The company wants to make the most money possible from selling pens.

(iii) Constraints:
Raw material requirement:
Pen A needs twice the raw material of pen B. If pen B needs \( x_2 \) units, then pen A needs \( 2x_1 \) units.
Total raw material available is 1000 pens per day.
\( \implies 2x_1 + x_2 \leq 1000 \)

Special clip requirement for pen A:
Only 400 clips are available for pen A per day.
\( \implies x_1 \leq 400 \)

Clip requirement for pen B:
Only 700 clips are available for pen B per day.
\( \implies x_2 \leq 700 \)

(iv) Non-negative restriction:
The number of pens produced cannot be negative.
\( \implies x_1 \geq 0, x_2 \geq 0 \)

Thus, the mathematical formulation of the Linear Programming Problem (LPP) is:
Maximize \( Z = 5x_1 + 3x_2 \)
Subject to the constraints:
\( 2x_1 + x_2 \leq 1000 \)
\( x_1 \leq 400 \)
\( x_2 \leq 700 \)
\( x_1, x_2 \geq 0 \)
In simple words: This problem asks us to find the best number of pens (A and B) to make to get the most profit. We have limits on raw materials and special clips, and we can't make a negative number of pens.

🎯 Exam Tip: When formulating LPPs, always clearly define variables, then state the objective function (what to maximize or minimize), and finally list all constraints, including non-negativity.

Question 2. A company produces two types of products say type A and B. Profits on the two types of product are Rs 30/- and 40/- per kg respectively. The data on resources required and availability of resources are given below.

Formulate this problem as a linear programming problem to maximize the profit.
Answer:
(i) Variables:
Let \( x_1 \) be the number of units of product A.
Let \( x_2 \) be the number of units of product B.

(ii) Objective function:
Profit on \( x_1 \) units of product A is \( 30x_1 \).
Profit on \( x_2 \) units of product B is \( 40x_2 \).
Total profit \( Z = 30x_1 + 40x_2 \).
The company wants to maximize profit, so the objective function is to maximize \( Z = 30x_1 + 40x_2 \). Making more profit is the main aim for any business.

(iii) Constraints:
Raw material constraint:
\( 60x_1 + 120x_2 \leq 12000 \)

Machining hours constraint:
\( 8x_1 + 5x_2 \leq 600 \)

Assembling hours constraint:
\( 3x_1 + 4x_2 \leq 500 \)

(iv) Non-negative constraints:
The number of products cannot be negative.
\( \implies x_1 \geq 0, x_2 \geq 0 \)

Thus, the mathematical formulation of the Linear Programming Problem (LPP) is:
Maximize \( Z = 30x_1 + 40x_2 \)
Subject to the constraints:
\( 60x_1 + 120x_2 \leq 12000 \)
\( 8x_1 + 5x_2 \leq 600 \)
\( 3x_1 + 4x_2 \leq 500 \)
\( x_1, x_2 \geq 0 \)
In simple words: We need to figure out how much of two products to make to get the most profit. There are limits on raw materials, machine time, and assembly time. We also can't make less than zero products.

🎯 Exam Tip: Pay close attention to the units (e.g., kgs, hours) and the "per unit" rates when writing out the objective function and constraints to avoid common errors.

Question 3. A company manufactures two models of voltage stabilizers viz., ordinary and autocut. All components of the stabilizers are purchased from outside sources, assembly and testing is carried out at company's own works. The assembly and testing time required for the two models are 0.8 hour each for ordinary and 1.20 hours each for auto-cut. Manufacturing capacity 720 hours at present is available per week. The market for the two models has been surveyed which suggests maximum weekly sale of 600 units of ordinary and 400 units of auto-cut. Profit per unit for ordinary and auto-cut models has been estimated at Rs 100 and Rs 150 respectively. Formulate the linear programming problem.
Answer:
(i) Variables:
Let \( x_1 \) be the number of ordinary voltage stabilizers.
Let \( x_2 \) be the number of auto-cut voltage stabilizers.

(ii) Objective function:
Profit on \( x_1 \) units of ordinary stabilizers is \( 100x_1 \).
Profit on \( x_2 \) units of auto-cut stabilizers is \( 150x_2 \).
Total profit \( Z = 100x_1 + 150x_2 \).
The company's aim is to maximize this profit, so the objective function is to maximize \( Z = 100x_1 + 150x_2 \). A higher profit means more success for the business.

(iii) Constraints:
Assembly and testing time constraint:
Time for \( x_1 \) ordinary units = \( 0.8x_1 \) hours.
Time for \( x_2 \) auto-cut units = \( 1.2x_2 \) hours.
Total manufacturing capacity is 720 hours per week.
\( \implies 0.8x_1 + 1.2x_2 \leq 720 \)

Maximum weekly sale for ordinary stabilizers:
\( \implies x_1 \leq 600 \)

Maximum weekly sale for auto-cut stabilizers:
\( \implies x_2 \leq 400 \)

(iv) Non-negative restrictions:
The number of stabilizers produced cannot be negative.
\( \implies x_1 \geq 0, x_2 \geq 0 \)

Thus, the mathematical formulation of the Linear Programming Problem (LPP) is:
Maximize \( Z = 100x_1 + 150x_2 \)
Subject to the constraints:
\( 0.8x_1 + 1.2x_2 \leq 720 \)
\( x_1 \leq 600 \)
\( x_2 \leq 400 \)
\( x_1, x_2 \geq 0 \)
In simple words: We need to set up a math problem to find the most profitable number of two types of stabilizers to make. We have limits on production time and how many of each type we can sell. We can't make negative amounts.

🎯 Exam Tip: When dealing with fractional time or resource requirements, ensure you multiply the variable by the correct decimal value for each unit, as shown with 0.8 and 1.2 hours here.

Question 4. Solve the following linear programming problems by graphical method.
(i) Maximize \( Z = 6x_1 + 8x_2 \) subject to constraints \( 30x_1 + 20x_2 \leq 300 \); \( 5x_1 + 10x_2 \leq 110 \); and \( x_1, x_2 \geq 0 \).
Answer:
(i) Given: Maximize \( Z = 6x_1 + 8x_2 \)
Constraints: \( 30x_1 + 20x_2 \leq 300 \)
\( 5x_1 + 10x_2 \leq 110 \)
\( x_1, x_2 \geq 0 \)

First, consider the inequality \( 30x_1 + 20x_2 \leq 300 \).
Convert to an equation: \( 30x_1 + 20x_2 = 300 \).
Divide by 10: \( 3x_1 + 2x_2 = 30 \).
Points for this line:

Next, consider the inequality \( 5x_1 + 10x_2 \leq 110 \).
Convert to an equation: \( 5x_1 + 10x_2 = 110 \).
Divide by 5: \( x_1 + 2x_2 = 22 \).
Points for this line:

To find the point of intersection (Point B) of the two lines:
\( 3x_1 + 2x_2 = 30 \) ....(1)
\( x_1 + 2x_2 = 22 \) ....(2)
Subtract equation (2) from equation (1):
\( (3x_1 + 2x_2) - (x_1 + 2x_2) = 30 - 22 \)
\( \implies 2x_1 = 8 \)
\( \implies x_1 = 4 \)
Substitute \( x_1 = 4 \) into equation (2):
\( 4 + 2x_2 = 22 \)
\( \implies 2x_2 = 18 \)
\( \implies x_2 = 9 \)
So, point B is \( (4, 9) \). This point is where the two constraint lines cross each other.

The feasible region (the area that satisfies all conditions) is OABC. This region is found by plotting the lines and shading the area that meets all inequalities, including \( x_1 \geq 0 \) and \( x_2 \geq 0 \).
The corner points of this feasible region are:
O \( (0, 0) \)
A \( (10, 0) \)
B \( (4, 9) \)
C \( (0, 11) \)

Now, we evaluate the objective function \( Z = 6x_1 + 8x_2 \) at each corner point to find the maximum value.

The maximum value of \( Z \) occurs at point B \( (4, 9) \).
Therefore, the optimal solution is \( x_1 = 4, x_2 = 9 \) and \( Z_{max} = 96 \). This means the highest profit is achieved when 4 units of \( x_1 \) and 9 units of \( x_2 \) are produced.
In simple words: To solve this, we drew lines for each rule (constraint) on a graph. Then we found the area where all the rules are met (feasible region). We checked the "profit" (objective function) at the corners of this area to find the highest profit. The best answer is to make 4 units of the first item and 9 units of the second item, giving a profit of 96.

🎯 Exam Tip: When using the graphical method, always clearly label your axes, plot points accurately, and shade the feasible region correctly. Remember to evaluate the objective function at all corner points of the feasible region to find the optimum.

Question 4.
(ii) Maximize \( Z = 22x_1 + 18x_2 \) subject to constraints \( 960x_1 + 640x_2 \leq 15360 \); \( x_1 + x_2 \leq 20 \) and \( x_1, x_2 \geq 0 \).
Answer:
(ii) Given: Maximize \( Z = 22x_1 + 18x_2 \)
Constraints: \( 960x_1 + 640x_2 \leq 15360 \)
\( x_1 + x_2 \leq 20 \)
\( x_1, x_2 \geq 0 \)

First, consider the inequality \( 960x_1 + 640x_2 \leq 15360 \).
Convert to an equation: \( 960x_1 + 640x_2 = 15360 \).
Divide by 320: \( 3x_1 + 2x_2 = 48 \).
Points for this line:

Next, consider the inequality \( x_1 + x_2 \leq 20 \).
Convert to an equation: \( x_1 + x_2 = 20 \).
Points for this line:

To find the point of intersection of the two lines:
\( 3x_1 + 2x_2 = 48 \) ....(1)
\( x_1 + x_2 = 20 \) ....(2)
Multiply equation (2) by -2 to eliminate \( x_2 \):
\( -2x_1 - 2x_2 = -40 \) ....(3)
Add equation (1) and equation (3):
\( (3x_1 + 2x_2) + (-2x_1 - 2x_2) = 48 + (-40) \)
\( \implies x_1 = 8 \)
Substitute \( x_1 = 8 \) into equation (2):
\( 8 + x_2 = 20 \)
\( \implies x_2 = 12 \)
So, the intersection point B is \( (8, 12) \). This is where the two constraint lines cross.

The feasible region satisfying all the given conditions is OABC.
The corner points of this feasible region are:
O \( (0, 0) \)
A \( (16, 0) \)
B \( (8, 12) \)
C \( (0, 20) \)

Now, we evaluate the objective function \( Z = 22x_1 + 18x_2 \) at each corner point.

The maximum value of \( Z \) occurs at point B \( (8, 12) \).
Therefore, the optimal solution is \( x_1 = 8, x_2 = 12 \) and \( Z_{max} = 392 \). This means the highest profit is achieved when 8 units of \( x_1 \) and 12 units of \( x_2 \) are produced.
In simple words: For this problem, we drew lines for the rules, found the area that fits all rules, and then tested the corners of this area. The biggest profit (392) came from making 8 units of the first item and 12 units of the second item.

🎯 Exam Tip: Simplify constraint equations before plotting by dividing by common factors (e.g., 960, 640, 15360 by 320) to work with smaller, easier numbers, reducing the chance of calculation errors.

Question 4.
(iii) Minimize \( Z = 3x_1 + 2x_2 \) subject to the constraints \( 5x_1 + x_2 \geq 10 \); \( x_1 + x_2 \geq 6 \); \( x_1 + 4x_2 \geq 12 \) and \( x_1, x_2 \geq 0 \).
Answer:
(iii) Given: Minimize \( Z = 3x_1 + 2x_2 \)
Constraints: \( 5x_1 + x_2 \geq 10 \)
\( x_1 + x_2 \geq 6 \)
\( x_1 + 4x_2 \geq 12 \)
\( x_1, x_2 \geq 0 \)

First, consider the inequality \( 5x_1 + x_2 \geq 10 \).
Convert to an equation: \( 5x_1 + x_2 = 10 \).
Points for this line:

Next, consider the inequality \( x_1 + x_2 \geq 6 \).
Convert to an equation: \( x_1 + x_2 = 6 \).
Points for this line:

Next, consider the inequality \( x_1 + 4x_2 \geq 12 \).
Convert to an equation: \( x_1 + 4x_2 = 12 \).
Points for this line:

To get point C (intersection of \( 5x_1 + x_2 = 10 \) and \( x_1 + x_2 = 6 \)):
\( 5x_1 + x_2 = 10 \) ....(1)
\( x_1 + x_2 = 6 \) ....(2)
Subtract equation (2) from equation (1):
\( 4x_1 = 4 \)
\( \implies x_1 = 1 \)
Substitute \( x_1 = 1 \) into equation (2):
\( 1 + x_2 = 6 \)
\( \implies x_2 = 5 \)
So, point C is \( (1, 5) \).

To get point B (intersection of \( x_1 + x_2 = 6 \) and \( x_1 + 4x_2 = 12 \)):
\( x_1 + x_2 = 6 \) ....(1)
\( x_1 + 4x_2 = 12 \) ....(2)
Subtract equation (2) from equation (1):
\( -3x_2 = -6 \)
\( \implies x_2 = 2 \)
Substitute \( x_2 = 2 \) into equation (1):
\( x_1 + 2 = 6 \)
\( \implies x_1 = 4 \)
So, point B is \( (4, 2) \).

The feasible region satisfying all the conditions is ABCD. This region is unbounded, meaning it extends infinitely in one direction, but the corner points help us find the minimum. The lines define the boundaries for this acceptable area.
The corner points of this feasible region are:
A \( (12, 0) \)
B \( (4, 2) \)
C \( (1, 5) \)
D \( (0, 10) \)

Now, we evaluate the objective function \( Z = 3x_1 + 2x_2 \) at each corner point.

The minimum value of \( Z \) occurs at point C \( (1, 5) \).
Therefore, the optimal solution is \( x_1 = 1, x_2 = 5 \) and \( Z_{min} = 13 \). This means the lowest cost or minimum value is achieved when 1 unit of \( x_1 \) and 5 units of \( x_2 \) are produced.
In simple words: For this problem, we drew lines for the rules, found the feasible region, and then checked the "cost" (objective function) at the corners of this region. We were looking for the lowest cost. The best answer is to make 1 unit of the first item and 5 units of the second item, giving a cost of 13.

🎯 Exam Tip: For minimization problems with "greater than or equal to" constraints, the feasible region is often unbounded. Always double-check that you've correctly identified the corner points of the bounded section of the feasible region, as the minimum will lie there.

Question 4.
(iv) Maximize \( Z = 40x_1 + 50x_2 \) subject to constraints \( 3x_1 + x_2 \leq 9 \); \( x_1 + 2x_2 \leq 8 \) and \( x_1, x_2 \geq 0 \).
Answer:
(iv) Given: Maximize \( Z = 40x_1 + 50x_2 \)
Constraints: \( 3x_1 + x_2 \leq 9 \)
\( x_1 + 2x_2 \leq 8 \)
\( x_1, x_2 \geq 0 \)

First, consider the inequality \( 3x_1 + x_2 \leq 9 \).
Convert to an equation: \( 3x_1 + x_2 = 9 \).
Points for this line:

Next, consider the inequality \( x_1 + 2x_2 \leq 8 \).
Convert to an equation: \( x_1 + 2x_2 = 8 \).
Points for this line:

To find the point of intersection of the two lines:
\( 3x_1 + x_2 = 9 \) ....(1)
\( x_1 + 2x_2 = 8 \) ....(2)
Multiply equation (1) by 2:
\( 6x_1 + 2x_2 = 18 \) ....(3)
Subtract equation (2) from equation (3):
\( (6x_1 + 2x_2) - (x_1 + 2x_2) = 18 - 8 \)
\( \implies 5x_1 = 10 \)
\( \implies x_1 = 2 \)
Substitute \( x_1 = 2 \) into equation (1):
\( 3(2) + x_2 = 9 \)
\( 6 + x_2 = 9 \)
\( \implies x_2 = 3 \)
So, the intersection point B is \( (2, 3) \). This is where the two constraint lines meet.

The feasible region satisfying all the conditions is OABC.
The corner points of this feasible region are:
O \( (0, 0) \)
A \( (3, 0) \)
B \( (2, 3) \)
C \( (0, 4) \)

Now, we evaluate the objective function \( Z = 40x_1 + 50x_2 \) at each corner point.

The maximum value of \( Z \) occurs at point B \( (2, 3) \).
Therefore, the optimal solution is \( x_1 = 2, x_2 = 3 \) and \( Z_{max} = 230 \). This shows that the highest profit is made when 2 units of \( x_1 \) and 3 units of \( x_2 \) are produced.
In simple words: We graphed the limits, found the common area, and then checked the profit at the corners. The highest profit (230) was found when we made 2 units of the first item and 3 units of the second item.

🎯 Exam Tip: Always make sure to check the origin \( (0,0) \) against the inequality to correctly determine which side of the line to shade for the feasible region, especially with "less than or equal to" constraints.

Question 4.
(v) Maximize \( Z = 20x_1 + 30x_2 \) subject to constraints \( 3x_1 + 3x_2 \leq 36 \); \( 5x_1 + 2x_2 \leq 50 \); \( 2x_1 + 6x_2 \leq 60 \) and \( x_1, x_2 \geq 0 \).
Answer:
(v) Given: Maximize \( Z = 20x_1 + 30x_2 \)
Constraints: \( 3x_1 + 3x_2 \leq 36 \)
\( 5x_1 + 2x_2 \leq 50 \)
\( 2x_1 + 6x_2 \leq 60 \)
\( x_1, x_2 \geq 0 \)

First, consider the inequality \( 3x_1 + 3x_2 \leq 36 \).
Convert to an equation: \( 3x_1 + 3x_2 = 36 \).
Divide by 3: \( x_1 + x_2 = 12 \).
Points for this line:

Next, consider the inequality \( 5x_1 + 2x_2 \leq 50 \).
Convert to an equation: \( 5x_1 + 2x_2 = 50 \).
Points for this line:

Next, consider the inequality \( 2x_1 + 6x_2 \leq 60 \).
Convert to an equation: \( 2x_1 + 6x_2 = 60 \).
Divide by 2: \( x_1 + 3x_2 = 30 \).
Points for this line:

To find the point of intersection C (between \( x_1 + x_2 = 12 \) and \( x_1 + 3x_2 = 30 \)):
\( x_1 + x_2 = 12 \) ....(1)
\( x_1 + 3x_2 = 30 \) ....(2)
Subtract equation (1) from equation (2):
\( (x_1 + 3x_2) - (x_1 + x_2) = 30 - 12 \)
\( \implies 2x_2 = 18 \)
\( \implies x_2 = 9 \)
Substitute \( x_2 = 9 \) into equation (1):
\( x_1 + 9 = 12 \)
\( \implies x_1 = 3 \)
So, point C is \( (3, 9) \).

To find the point of intersection B (between \( x_1 + x_2 = 12 \) and \( 5x_1 + 2x_2 = 50 \)):
\( x_1 + x_2 = 12 \) ....(1)
\( 5x_1 + 2x_2 = 50 \) ....(2)
Multiply equation (1) by 2:
\( 2x_1 + 2x_2 = 24 \) ....(3)
Subtract equation (3) from equation (2):
\( (5x_1 + 2x_2) - (2x_1 + 2x_2) = 50 - 24 \)
\( \implies 3x_1 = 26 \)
\( \implies x_1 = \frac{26}{3} \approx 8.67 \)
Substitute \( x_1 = \frac{26}{3} \) into equation (1):
\( \frac{26}{3} + x_2 = 12 \)
\( x_2 = 12 - \frac{26}{3} \)
\( x_2 = \frac{36 - 26}{3} = \frac{10}{3} \approx 3.33 \)
So, point B is \( (\frac{26}{3}, \frac{10}{3}) \).

The feasible region satisfying all the conditions is OABCD. This region is the area where all the given inequalities are true.
The corner points of this feasible region are:
O \( (0, 0) \)
A \( (10, 0) \)
B \( (\frac{26}{3}, \frac{10}{3}) \)
C \( (3, 9) \)
D \( (0, 10) \)

Now, we evaluate the objective function \( Z = 20x_1 + 30x_2 \) at each corner point.

The maximum value of \( Z \) occurs at point C \( (3, 9) \).
Therefore, the optimal solution is \( x_1 = 3, x_2 = 9 \) and \( Z_{max} = 330 \). This shows the most profit is made when 3 units of \( x_1 \) and 9 units of \( x_2 \) are produced.
In simple words: We found the best production mix (3 of the first item, 9 of the second) by graphing the limits, finding the common region, and checking the profit at each corner point. This gives the highest profit of 330.

🎯 Exam Tip: When corner points involve fractions, it's crucial to use exact fractional values in calculations for the objective function to maintain precision, rather than rounding too early. Round only the final result if necessary.

Question 4. (vi) Minimize \( Z = 20x_1 + 40x_2 \) subject to the constraints \( 36x_1 + 6x_2 \ge 108 \); \( 3x_1 + 12x_2 \ge 36 \); \( 20x_1 + 10x_2 \ge 100 \) and \( x_1, x_2 \ge 0 \).
Answer: First, let's find the points for each constraint by treating them as equations. This helps us draw the lines on a graph and identify the feasible region.

For the first constraint, \( 36x_1 + 6x_2 \ge 108 \):
Let \( 36x_1 + 6x_2 = 108 \).
Dividing by 6, we get \( 6x_1 + x_2 = 18 \).

For the second constraint, \( 3x_1 + 12x_2 \ge 36 \):
Let \( 3x_1 + 12x_2 = 36 \).
Dividing by 3, we get \( x_1 + 4x_2 = 12 \).

For the third constraint, \( 20x_1 + 10x_2 \ge 100 \):
Let \( 20x_1 + 10x_2 = 100 \).
Dividing by 10, we get \( 2x_1 + x_2 = 10 \).

The problem also includes non-negative restrictions: \( x_1 \ge 0 \) and \( x_2 \ge 0 \). These ensure our solution stays in the first quadrant of the graph. The feasible region is the area on the graph that satisfies all these conditions. This region is unbounded, as seen in the graph below, and its corner points are where these lines intersect.

The feasible region that meets all the given conditions is the unbounded area shown in the graph. The important corner points of this region, which help determine the minimum value, are A(12, 0), B(4, 2), C(2, 6), and D(0, 18). We now evaluate the objective function \( Z = 20x_1 + 40x_2 \) at each of these corner points.

From the calculations, the minimum value of \( Z \) is 160, which occurs at the corner point B(4, 2). This means to achieve the lowest possible value for \( Z \), \( x_1 \) should be 4 and \( x_2 \) should be 2. Linear programming helps businesses find the most efficient way to use their resources by minimizing costs or maximizing profits under given conditions.
In simple words: We find the points for each rule and draw lines on a graph. The area where all rules are true is the "feasible region." We then check the corners of this region to find the lowest value for Z. The lowest value of Z, which is 160, happens when \( x_1 \) is 4 and \( x_2 \) is 2.

🎯 Exam Tip: For minimization problems with "greater than or equal to" constraints, the feasible region is often unbounded, extending outwards. Always check the corner points of the shaded region where the lines intersect to find the minimum value of the objective function.