GSEB Class 12 Maths Solutions Chapter 12 સુરેખ આયોજન Exercise 12.2

Gujarat Board Textbook Solutions Class 12 Chapter 12 Linear Programming Ex 12.2

 

Question 1. Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contains atleast 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs.60/kg and Food Q costs Rs.80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B, while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
Answer: Let \( x \) kg of food P and \( y \) kg of food Q be combined. We are given the following information:

The amount of vitamin A in \( x \) kg of food P and \( y \) kg of food Q is \( 3x + 4y \). The minimum vitamin A needed is 8 units.
\( \implies 3x + 4y \geq 8 \)
The amount of vitamin B in \( x \) kg of food P and \( y \) kg of food Q is \( 5x + 2y \). The minimum vitamin B needed is 11 units.
\( \implies 5x + 2y \geq 11 \)
The cost of \( x \) kg of food P and \( y \) kg of food Q is \( 60x + 80y \). The objective function \( Z \) is \( 60x + 80y \). Constraints are \( 3x + 4y \geq 8 \), \( 5x + 2y \geq 11 \), and \( x, y > 0 \).
(i) The line \( 3x + 4y = 8 \) goes through points A\( ( \frac{8}{3}, 0 ) \) and B\( (0, 2) \). If we put \( x = 0, y = 0 \) into \( 3x + 4y \geq 8 \), we get \( 0 \geq 8 \), which is incorrect.
\( \implies \) The area shown by \( 3x + 4y \geq 8 \) is on and above the line AB.
(ii) The line \( 5x + 2y = 11 \) goes through points C\( ( \frac{11}{5}, 0 ) \) and D\( (0, \frac{11}{2}) \). If we put \( x = 0, y = 0 \) into \( 5x + 2y \geq 11 \), we get \( 0 \geq 11 \), which is incorrect.
\( \implies \) The area \( 5x + 2y \geq 11 \) is on and above the line CD.
(iii) \( x \geq 0 \) is the area to the right of the y-axis and on it.
(iv) \( y \geq 0 \) is the area above the x-axis and on it. The shaded area, YDPAX, represents the feasible region. P is the point where AB and CD intersect. These lines are: 1. \( 3x + 4y = 8 \) 2. \( 5x + 2y = 11 \) To find the intersection point, we multiply equation (2) by 2 and then subtract equation (1) from it: \( 2(5x + 2y) - (3x + 4y) = 2(11) - 8 \) \( 10x + 4y - 3x - 4y = 22 - 8 \) \( 7x = 14 \) \( x = 2 \) Now, substitute \( x = 2 \) into equation (1): \( 3(2) + 4y = 8 \) \( 6 + 4y = 8 \) \( 4y = 2 \) \( y = \frac{1}{2} \)
\( \implies \) The point P is \( (2, \frac{1}{2}) \). Now, we calculate \( Z = 60x + 80y \) at the corner points: At D\( (0, \frac{11}{2}) \), \( Z = 80 \times \frac{11}{2} = 440 \) At P\( (2, \frac{1}{2}) \), \( Z = 60 \times 2 + 80 \times \frac{1}{2} = 120 + 40 = 160 \) At A\( ( \frac{8}{3}, 0 ) \), \( Z = 60 \times \frac{8}{3} + 0 = 160 \) The minimum value of \( Z \) is 160. However, the feasible region is unbounded. So, we consider the inequality \( 60x + 80y < 160 \), which simplifies to \( 3x + 4y < 8 \). This shows that this area is below the line AD: \( 3x + 4y = 8 \).
\( \implies \) There is no common point between the feasible region and \( 3x + 4y < 8 \). Therefore, the lowest cost \( Z \) is Rs.160 at all the points found on the line segment that joins A\( ( \frac{8}{3}, 0 ) \) and P\( (2, \frac{1}{2}) \).
In simple words: To find the lowest cost for the food mixture, we set up equations based on vitamin needs and food costs. We find where the lines cross on a graph and check the cost at those points. The lowest cost will be Rs.160.

Exam Tip: For unbounded feasible regions, always test the inequality \( Z < \text{minimum value} \) to confirm if the minimum exists and where it occurs.

 

Question 2. One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes, which can be made from 5 kg of flour and 1 kg of fat, assuming that there is no shortage of the other ingradients, used in making the cakes.
Answer: Let \( x \) be the number of cakes of the first type and \( y \) be the number of cakes of the second type. We have the following data:

The objective function is \( Z = x + y \), which we want to maximise. The highest amount of flour used is 5000 g (which is 5 kg).
\( \implies 200x + 100y \leq 5000 \) This can be simplified to:
\( 2x + y \leq 50 \) (1) The highest amount of fat used is 1000 g (which is 1 kg).
\( \implies 25x + 50y \leq 1000 \) This can be simplified to:
\( x + 2y \leq 40 \) (2) Also, the number of cakes cannot be negative:
\( x, y \geq 0 \) (3)
(i) The line \( 2x + y = 50 \) passes through the points A\( (25, 0) \) and B\( (0, 50) \). If we substitute \( x = 0, y = 0 \) into \( 2x + y \leq 50 \), we get \( 0 \leq 50 \), which is correct.
\( \implies \) The region represents the area on and below line AB.
(ii) The line \( x + 2y = 40 \) passes through C\( (40, 0) \) and D\( (0, 20) \). If we substitute \( x = 0, y = 0 \) into \( x + 2y \leq 40 \), we get \( 0 \leq 40 \), which is correct.
\( \implies \) The region represents the area on and below line CD.
(iii) \( x \geq 0 \) is the region on the y-axis and to its right.
(iv) \( y \geq 0 \) is the region on the x-axis and above it. The feasible region is OAPD, which is a bounded region shown by the shaded area in the figure (omitted). The lines AB \( 2x + y = 50 \) and CD \( x + 2y = 40 \) intersect at point P. To find the intersection point, we solve these equations: 1. \( 2x + y = 50 \) 2. \( x + 2y = 40 \) Multiply equation (1) by 2: \( 4x + 2y = 100 \) Subtract equation (2) from this new equation: \( (4x + 2y) - (x + 2y) = 100 - 40 \) \( 3x = 60 \) \( x = 20 \) Substitute \( x = 20 \) into equation (1): \( 2(20) + y = 50 \) \( 40 + y = 50 \) \( y = 10 \)
\( \implies \) The point P is \( (20, 10) \). Now, we calculate \( Z = x + y \) at the corner points of the feasible region: At A\( (25, 0) \), \( Z = 25 + 0 = 25 \) At P\( (20, 10) \), \( Z = 20 + 10 = 30 \) At D\( (0, 20) \), \( Z = 0 + 20 = 20 \) At O\( (0, 0) \), \( Z = 0 \) Thus, the maximum value of \( Z \) is 30, which happens at P\( (20, 10) \). This means that 20 cakes of the first type and 10 cakes of the second type will give the highest number of cakes.
In simple words: To make the most cakes with the limited flour and fat, the factory should produce 20 cakes of type I and 10 cakes of type II. This will make a total of 30 cakes.

Exam Tip: Remember to convert all quantities to the same units (e.g., grams for flour and fat) before setting up the inequalities.

 

Question 3. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of a and 3 hours of craftman's time in its making, while a cricket bat takes 3 hours of machine time and 1 hour of craftman's time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time.
(i) What number of rackets and bats must be made, if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat are Rs.20 and Rs.10 respectively, find the maximum profit to the factory, when it works at full capacity.
Answer: Let \( x \) represent the number of tennis rackets and \( y \) represent the number of cricket bats produced in one day by the factory. We have the following information:

The total machine hours used are \( 1.5x + 3y \). The highest machine time available is 42 hours.
\( \implies 1.5x + 3y \leq 42 \) Multiply by 2 to clear decimals:
\( 3x + 6y \leq 84 \) Divide by 3:
\( x + 2y \leq 28 \) (1) The total craftsman's hours used are \( 3x + y \). The highest craftsman's time available is 24 hours.
\( \implies 3x + y \leq 24 \) (2) Also, the number of rackets and bats cannot be negative:
\( x \geq 0, y \geq 0 \)
(a) The line \( x + 2y = 28 \) goes through A\( (28, 0) \) and B\( (0, 14) \). If we substitute \( x = 0, y = 0 \) into \( x + 2y \leq 28 \), we get \( 0 \leq 28 \), which is correct.
\( \implies x + 2y \leq 28 \) represents the area on and below AB.
(b) The line \( 3x + y = 24 \) goes through C\( (8, 0) \) and D\( (0, 24) \). If we substitute \( x = 0, y = 0 \) into \( 3x + y \leq 24 \), we get \( 0 \leq 24 \), which is correct.
\( \implies 3x + y \leq 24 \) represents the area on and below CD.
(c) \( x \geq 0 \) represents the area on and to the right of the y-axis.
(d) \( y \geq 0 \) represents the area on and above the x-axis. The shaded area BPCO (in the figure, which is omitted) is the feasible region. The two lines AB and CD are: 1. \( x + 2y = 28 \) 2. \( 3x + y = 24 \) To find the intersection point P, we multiply equation (2) by 2 and subtract equation (1) from it: \( 2(3x + y) - (x + 2y) = 2(24) - 28 \) \( 6x + 2y - x - 2y = 48 - 28 \) \( 5x = 20 \) \( x = 4 \) Substitute \( x = 4 \) into equation (2): \( 3(4) + y = 24 \) \( 12 + y = 24 \) \( y = 12 \) Thus, these lines meet at P\( (4, 12) \).
(i) We need to find the number of rackets and bats if the factory works at full capacity. This means maximizing \( Z = x + y \). At B\( (0, 14) \), \( Z = 0 + 14 = 14 \) At P\( (4, 12) \), \( Z = 4 + 12 = 16 \) At C\( (8, 0) \), \( Z = 8 + 0 = 8 \) The highest value of \( Z \) is 16. This happens when 4 tennis rackets and 12 cricket bats are made, so the factory operates at its greatest capacity.
(ii) The profit function is \( Z = 20x + 10y \). Now we calculate the profit at the corner points: At B\( (0, 14) \), \( Z = 0 + 10 \times 14 = 140 \) At P\( (4, 12) \), \( Z = 20 \times 4 + 10 \times 12 = 80 + 120 = 200 \) At C\( (8, 0) \), \( Z = 20 \times 8 + 0 = 160 \) At D\( (0, 0) \), \( Z = 0 \) Therefore, the highest profit is Rs.200. This occurs when 4 tennis rackets and 12 cricket bats are produced.
In simple words: For the factory to work at its maximum, it should make 4 tennis rackets and 12 cricket bats. This will give the biggest possible profit of Rs.200.

Exam Tip: Clearly distinguish between total items (capacity) and profit calculations, and always evaluate the objective function at all corner points of the feasible region.

 

Question 4. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs.17.50 per packages on nuts and Rs.7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?
Answer: Let \( x \) be the number of packages of nuts and \( y \) be the number of packages of bolts produced. We have the following data:

Machine A is utilized for \( x \times 1 + y \times 3 \) hours. The highest time available for Machine A is 12 hours.
\( \implies x + 3y \leq 12 \) Machine B is utilized for \( 3 \times x + 1 \times y \) hours. The highest time available for Machine B is 12 hours.
\( \implies 3x + y \leq 12 \) The profit function is \( Z = (17.50)x + (7.00)y \). Thus, the objective function is \( Z = 17.5x + 7y \). We want to maximize this subject to the constraints: \( x + 3y \leq 12 \), \( 3x + y \leq 12 \), and \( x, y \geq 0 \).
(i) The line \( x + 3y = 12 \) goes through A\( (12, 0) \) and B\( (0, 4) \). If we substitute \( x = 0, y = 0 \) into \( x + 3y \leq 12 \), we get \( 0 \leq 12 \), which is correct.
\( \implies x + 3y \leq 12 \) is on and below AB.
(ii) The line \( 3x + y = 12 \) goes through C\( (4, 0) \) and D\( (0, 12) \). If we substitute \( x = 0, y = 0 \) into \( 3x + y \leq 12 \), we get \( 0 \leq 12 \), which is correct.
\( \implies 3x + y \leq 12 \) is on and below CD.
(iii) \( x \geq 0 \) is the area to the right of the y-axis and on it.
(iv) \( y \geq 0 \) is the area above the x-axis and on it. The shaded area BPCO (in the figure, which is omitted) is the feasible region. The point P is the intersection of the lines: 1. AB: \( x + 3y = 12 \) 2. CD: \( 3x + y = 12 \) To find the intersection point P, we multiply equation (1) by 3 and subtract equation (2) from it: \( 3(x + 3y) - (3x + y) = 3(12) - 12 \) \( 3x + 9y - 3x - y = 36 - 12 \) \( 8y = 24 \) \( y = 3 \) Substitute \( y = 3 \) into equation (1): \( x + 3(3) = 12 \) \( x + 9 = 12 \) \( x = 3 \)
\( \implies \) The point P is \( (3, 3) \). Now, we calculate \( Z = 17.5x + 7y \) at the corner points: At B\( (0, 4) \), \( Z = 0 + 7 \times 4 = 28 \) At P\( (3, 3) \), \( Z = 17.5 \times 3 + 7 \times 3 = 52.5 + 21 = 73.5 \) At C\( (4, 0) \), \( Z = 17.5 \times 4 + 0 = 70 \) At O\( (0, 0) \), \( Z = 0 \) The highest profit is Rs.73.50. This occurs when 3 packages of nuts and 3 packages of bolts are produced.
In simple words: To get the most profit, the manufacturer should make 3 packages of nuts and 3 packages of bolts. This will give a profit of Rs.73.50.

Exam Tip: Clearly label your constraints and objective function. Remember to check all corner points of the feasible region to find the maximum or minimum value.

 

Question 5. A factory manufactures two types of screws A and B, each type requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machine to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machine to manufacture a package of screws B. Each machine is available for at the most 4 hours on airy day. The manufacturer can sell a package of screws A at a profit of Rs.7 and screws B at a profit of Rs.10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Also, determine the miximum profit.
Answer: Let the manufacturer produce \( x \) packages of screws A and \( y \) packages of screws B. The time taken for \( x \) packages of screws A and \( y \) packages of screws B on the automatic machine is \( (4x + 6y) \) minutes. The time taken on the hand-operated machine is \( (6x + 3y) \) minutes. Each machine is available for a maximum of 4 hours, which is \( 4 \times 60 = 240 \) minutes. So, we have the following constraints: For the automatic machine: \( 4x + 6y \leq 240 \) Divide by 2: \( 2x + 3y \leq 120 \) For the hand-operated machine: \( 6x + 3y \leq 240 \) Divide by 3: \( 2x + y \leq 80 \) The profit from selling \( x \) packages of screws A and \( y \) packages of screws B is \( Z = 7x + 10y \). Our problem is to find \( x \) and \( y \) such that \( Z = 7x + 10y \) is maximized, subject to the constraints: \( 2x + 3y \leq 120 \) \( 2x + y \leq 80 \) \( x \geq 0 \) and \( y \geq 0 \) We will plot the graphs for the equations \( 2x + 3y = 120 \) and \( 2x + y = 80 \). The feasible region of the graph that satisfies the inequalities \( 2x + 3y \leq 120 \) and \( 2x + y \leq 80 \) is OABC, as shown shaded in the figure (omitted). The coordinates of O, A, B, and C are O\( (0, 0) \), A\( (0, 40) \), B\( (30, 20) \), and C\( (40, 0) \). We will find the values of \( Z \) at these corner points: At A\( (0, 40) \), \( Z = 7x + 10y = 0 + 10 \times 40 = 400 \) At B\( (30, 20) \), \( Z = 7 \times 30 + 10 \times 20 = 210 + 200 = 410 \) At C\( (40, 0) \), \( Z = 7 \times 40 + 0 = 280 \) At O\( (0, 0) \), \( Z = 0 \) Therefore, the highest profit is Rs.410, obtained when 30 screws of type A and 20 screws of type B are produced.
In simple words: To make the most profit, the factory owner should produce 30 packages of screw type A and 20 packages of screw type B. This will result in the highest profit of Rs.410.

Exam Tip: Convert all time units to be consistent (e.g., minutes) before setting up inequalities to avoid errors in calculations.

 

Question 6. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp, while if takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/ cutting machine for at the most 12 hours. The profit on the sale of a lamp is Rs.5 and that from a shade is Rs.3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?
Answer: Let the manufacturer produce \( x \) pedestal lamps and \( y \) wooden shades. The time taken for \( x \) pedestal lamps and \( y \) wooden shades on the grinding/cutting machine is \( (2x + y) \) hours. The time taken for \( x \) pedestal lamps and \( y \) shades on the sprayer is \( (3x + 2y) \) hours. The grinding/cutting machine is available for a maximum of 12 hours, so: \( 2x + y \leq 12 \) The sprayer is available for a maximum of 20 hours, so: \( 3x + 2y \leq 20 \) The profit from the sale of \( x \) lamps and \( y \) shades is \( Z = 5x + 3y \). Our problem is to maximize \( Z = 5x + 3y \) subject to the constraints: \( 3x + 2y \leq 20 \) \( 2x + y \leq 12 \) \( x \geq 0 \), \( y \geq 0 \) We will plot the equations \( 3x + 2y = 20 \) and \( 2x + y = 12 \) on a graph (omitted).
(i) The line \( 2x + y = 12 \) passes through A\( (6, 0) \) and B\( (0, 12) \). If we substitute \( x = 0, y = 0 \) into \( 2x + y \leq 12 \), we get \( 0 \leq 12 \), which is correct.
\( \implies 2x + y \leq 12 \) lies on and below AB.
(ii) The line \( 3x + 2y = 20 \) passes through C\( ( \frac{20}{3}, 0 ) \) and D\( (0, 10) \). If we substitute \( x = 0, y = 0 \) into \( 3x + 2y \leq 20 \), we get \( 0 \leq 20 \), which is correct.
\( \implies 3x + 2y \leq 20 \) lies on and below CD.
(iii) \( x \geq 0 \) is the region on and to the right of the y-axis.
(iv) \( y \geq 0 \) is the region on and above the x-axis. The shaded area OAPD (in the figure, which is omitted) is the feasible region, where P is the point of intersection of AB and CD. The lines are: 1. AB: \( 2x + y = 12 \) 2. CD: \( 3x + 2y = 20 \) To find the intersection point P, we multiply equation (1) by 2 and subtract equation (2) from it: \( 2(2x + y) - (3x + 2y) = 2(12) - 20 \) \( 4x + 2y - 3x - 2y = 24 - 20 \) \( x = 4 \) Substitute \( x = 4 \) into equation (1): \( 2(4) + y = 12 \) \( 8 + y = 12 \) \( y = 4 \)
\( \implies \) The point P is \( (4, 4) \). Now, we calculate \( Z = 5x + 3y \) at the corner points: At A\( (6, 0) \), \( Z = 30 \) At P\( (4, 4) \), \( Z = 5(4) + 3(4) = 20 + 12 = 32 \) At D\( (0, 10) \), \( Z = 0 + 3(10) = 30 \) At O\( (0, 0) \), \( Z = 0 \) Thus, the manufacturer should produce 4 lamps and 4 shades to get a maximum profit of Rs.32.
In simple words: To get the highest profit, the manufacturer should make 4 pedestal lamps and 4 wooden shades each day. This will result in a maximum profit of Rs.32.

Exam Tip: Graphing the feasible region clearly helps in identifying all corner points, which are critical for evaluating the objective function.

 

Question 7. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profits are Rs.5 each for type A and Rs.6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?
Answer: Let the company manufacture \( x \) souvenirs of type A and \( y \) souvenirs of type B. The time taken for cutting \( x \) souvenirs of type A and \( y \) souvenirs of type B is \( (5x + 8y) \) minutes. The time taken for assembling \( x \) souvenirs of type A and \( y \) souvenirs of type B is \( (10x + 8y) \) minutes. Since 3 hours 20 minutes are available for cutting, this equals \( (3 \times 60) + 20 = 180 + 20 = 200 \) minutes. So, we have the constraint for cutting: \( 5x + 8y \leq 200 \) Since 4 hours are available for assembling, this equals \( 4 \times 60 = 240 \) minutes. So, we have the constraint for assembling: \( 10x + 8y \leq 240 \) Divide by 2: \( 5x + 4y \leq 120 \) The profit for souvenir type A is Rs.5 and for type B is Rs.6. Thus, our Linear Programming Problem (LPP) is to maximize profit \( Z = 5x + 6y \), subject to the constraints: \( 5x + 8y \leq 200 \) \( 5x + 4y \leq 120 \) \( x \geq 0 \), \( y \geq 0 \) We represent the equations \( 5x + 8y = 200 \) and \( 5x + 4y = 120 \) graphically (omitted). For this, we can use the following table to find points for plotting the lines: For \( 5x + 4y = 120 \):

For \( 5x + 8y = 200 \):The part of the graph that satisfies the inequalities \( 5x + 8y \leq 200 \) and \( 5x + 4y \leq 120 \) is OABC, which is shown as the shaded area in the figure (omitted). The coordinates of points O, A, B, and C are O\( (0, 0) \), A\( (24, 0) \), B\( (8, 20) \), and C\( (0, 25) \). The objective function is \( Z = 5x + 6y \). Now we calculate \( Z \) at the corner points: At A\( (24, 0) \), \( Z = 5 \times 24 + 0 = 120 \) At B\( (8, 20) \), \( Z = 5 \times 8 + 6 \times 20 = 40 + 120 = 160 \) At C\( (0, 25) \), \( Z = 0 + 6 \times 25 = 150 \) At D\( (0, 0) \), \( Z = 0 \) Thus, the highest profit is Rs.160, which happens when 8 souvenirs of type A and 20 souvenirs of type B are manufactured.
In simple words: To achieve the greatest profit, the company should make 8 souvenirs of type A and 20 souvenirs of type B. This will give them a maximum profit of Rs.160.

Exam Tip: Remember to convert all time units to a common base (e.g., minutes) before formulating constraints to ensure consistency.

 

Question 8. A merchant plans to sell two types of personal computers a desk top model and a portable model that will cost Rs.25000 and Rs.40000 respectively. The total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get a maximum profit, if he does not want to invest more than Rs.70 lakh and, if his profit on the desktop model is Rs.4500 and the portable is Rs.5000.
Answer: Let \( x \) be the number of desktop computers and \( y \) be the number of portable computers. The total monthly demand for computers does not go over 250 units.
\( \implies x + y \leq 250 \) The cost of one desktop computer is Rs.25000 and one portable computer is Rs.40000. The cost of \( x \) desktop and \( y \) portable computers is \( (25000x + 40000y) \). The highest investment allowed is Rs.70 lakhs, which is Rs.70,00,000.
\( \implies 25000x + 40000y \leq 7000000 \) Divide by 5000:
\( 5x + 8y \leq 1400 \) The profit on one desktop computer is Rs.4500 and on one portable computer is Rs.5000. The total profit \( Z = 4500x + 5000y \). Therefore, the objective function is to maximize \( Z = 4500x + 5000y \), and the constraints are \( x + y \leq 250 \), \( 5x + 8y \leq 1400 \), and \( x, y \geq 0 \).
(i) The line \( x + y = 250 \) passes through A\( (250, 0) \) and B\( (0, 250) \). If we substitute \( x = 0, y = 0 \) into \( x + y \leq 250 \), we get \( 0 \leq 250 \), which is correct.
\( \implies x + y \leq 250 \) lies on and below AB.
(ii) The line \( 5x + 8y = 1400 \) passes through C\( (280, 0) \) and D\( (0, 175) \). If we substitute \( x = 0, y = 0 \) into \( 5x + 8y \leq 1400 \), we get \( 0 \leq 1400 \), which is correct.
\( \implies 5x + 8y \leq 1400 \) lies on and below CD.
(iii) \( x \geq 0 \) lies on and to the right of the y-axis.
(iv) \( y \geq 0 \) lies on and above the x-axis. The shaded area OAPD (in the figure, which is omitted) represents the feasible region, where P is the point where AB and CD intersect. The equations for the lines are: 1. \( 5x + 8y = 1400 \) 2. \( x + y = 250 \) To find the intersection point P, we multiply equation (2) by 8 and subtract equation (1) from it: \( 8(x + y) - (5x + 8y) = 8(250) - 1400 \) \( 8x + 8y - 5x - 8y = 2000 - 1400 \) \( 3x = 600 \) \( x = 200 \) Substitute \( x = 200 \) into equation (2): \( 200 + y = 250 \) \( y = 50 \) The point P is \( (200, 50) \). Now we calculate the values of \( Z \) at the corner points A, P, D, O: At A\( (250, 0) \), \( Z = 4500x + 5000y = 4500 \times 250 + 0 = 1125000 \) At P\( (200, 50) \), \( Z = 4500 \times 200 + 5000 \times 50 = 900000 + 250000 = 1150000 \) At D\( (0, 175) \), \( Z = 0 + 5000 \times 175 = 875000 \) At O\( (0, 0) \), \( Z = 0 \)
\( \implies \) The highest profit of Rs.11,50,000 is achieved when the merchant stocks 200 desktop computers and 50 portable computers.
In simple words: To get the most profit, the merchant should stock 200 desktop computers and 50 portable computers. This will give a profit of Rs.11,50,000.

Exam Tip: Be careful with large numbers and unit conversions (e.g., lakhs to standard units) when setting up the initial constraints.

 

Question 8. Determine the number of units of each type of computers which the merchant should stock to get a maximum profit, if he does not want to invest more than ₹70 lakh and, if his profit on the desktop model is ₹4500 and the portable is ₹5000.
Answer: The maximum profit of Rs. 11,50,000 is obtained when the merchant stocks 200 desktop computers and 50 portable computers. This strategy helps to achieve the highest possible earnings within the given investment limits.
In simple words: To make the most money (Rs. 11,50,000), the seller should keep 200 desktop computers and 50 portable computers in stock.

Exam Tip: Remember to express all monetary values consistently, e.g., converting lakhs to rupees early in the problem to avoid calculation errors.

 

Question 9. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F\( _1 \) costs ₹4 per unit food and F\( _2 \) costs ₹6 per unit. A unit of food F\( _1 \) contains at least 3 units of vitamin A and 4 units of minerals. One unit of food F\( _2 \) contains at least 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimum nutritional requirements.
Answer: Let \( x \) units of food F\( _1 \) and \( y \) units of food F\( _2 \) be required. The given information is presented in the table below:

The total quantity of vitamin A in \( x \) kg of food P and \( y \) kg of food Q is \( 3x + 6y \). The least quantity of vitamin A required is 80 units.
\( \implies 3x + 6y \ge 80 \) (A) The total quantity of vitamin B in \( x \) kg of food P and \( y \) kg of food Q is \( 4x + 3y \). The least quantity of minerals required is 100 units.
\( \implies 4x + 3y \ge 100 \) (B) Therefore, the Linear Programming Problem (LPP) is: To minimise \( Z = 4x + 6y \). Constraints are \( 3x + 6y \ge 80 \), \( 4x + 3y \ge 100 \) and \( x, y \ge 0 \). (i) The line \( 3x + 6y = 80 \) passes through \( A(\frac{80}{3}, 0) \) and \( B(0, \frac{40}{3}) \). When \( x = 0, y = 0 \) is put into \( 3x + 6y \ge 80 \), we get \( 0 \ge 80 \), which is false.
\( \implies \) The region represented by \( 3x + 6y \ge 80 \) lies on and above the line AB. (ii) The line \( 4x + 3y = 100 \) passes through \( C(25, 0) \) and \( D(0, \frac{100}{3}) \). When \( x = 0, y = 0 \) is put into \( 4x + 3y \ge 100 \), we get \( 0 \ge 100 \), which is false.
\( \implies \) The region represented by \( 4x + 3y \ge 100 \) lies on and above the line CD. (iii) \( x \ge 0 \) represents the region on and to the right of the y-axis. (iv) \( y \ge 0 \) represents the region on and above the x-axis. The shaded area YDPAX is the feasible region, which is an unbounded region, where P is the point of intersection of AB and CD. The lines are: \( 3x + 6y = 80 \) .......... (1) \( 4x + 3y = 100 \) .......... (2) Multiplying equation (2) by 2 and subtracting (1) from it, we get: \( 8x + 6y - (3x + 6y) = 200 - 80 \) \( \implies 5x = 120 \) \( \implies x = 24 \) From (1), \( 3(24) + 6y = 80 \) \( 72 + 6y = 80 \) \( 6y = 80 - 72 \) \( 6y = 8 \) \( \implies y = \frac{8}{6} = \frac{4}{3} \) So, the point P is \( (24, \frac{4}{3}) \). Values of Z at the corner points: At D\( (0, \frac{100}{3}) \), \( Z = 4(0) + 6(\frac{100}{3}) = 0 + 200 = 200 \) At P\( (24, \frac{4}{3}) \), \( Z = 4(24) + 6(\frac{4}{3}) = 96 + 8 = 104 \) At A\( (\frac{80}{3}, 0) \), \( Z = 4(\frac{80}{3}) + 6(0) = \frac{320}{3} = 106\frac{2}{3} \) The minimum value of Z is 104. Since the feasible region is unbounded, we also consider the inequality \( 4x + 6y < 104 \) or \( 2x + 3y < 52 \). The line \( 2x + 3y = 52 \) passes through \( (26, 0) \) and \( (0, 17\frac{1}{3}) \). There is no point common between the feasible region and this inequality. Therefore, the minimum cost of the diet is Rs. 104.In simple words: To make sure the diet has enough vitamin A and minerals while costing the least, we need to find the right mix of food F1 and F2. The lowest cost we can get is Rs. 104.

Exam Tip: When dealing with unbounded feasible regions in minimization problems, always remember to check if the objective function can take values less than the minimum value found at the corner points by considering the inequality \( Z < Z_{\text{min}} \).

 

Question 10. There are two types of fertilisers F\( _1 \) and F\( _2 \). F\( _1 \) contains 10% nitrogen and 6% phosphoric acid and F\( _2 \) consists of 5% of nitrogen and 10% of phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F\( _1 \) costs ₹6/kg and F\( _2 \) costs ₹5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Answer: Let \( x \) kg of fertiliser F\( _1 \) and \( y \) kg of fertiliser F\( _2 \) be required. We are given the following information:

The quantity of nitrogen from F\( _1 \) and F\( _2 \) is \( 10\% \) of \( x + 5\% \) of \( y = \frac{x}{10} + \frac{y}{20} \). The least quantity of nitrogen required is 14 kg.
\( \implies \frac{x}{10} + \frac{y}{20} \ge 14 \) or \( 2x + y \ge 280 \). The quantity of phosphoric acid from F\( _1 \) and F\( _2 \) is \( 6\% \) of \( x + 10\% \) of \( y = \frac{6x}{100} + \frac{y}{10} \). The least quantity of phosphoric acid required is 14 kg.
\( \implies \frac{6x}{100} + \frac{y}{10} \ge 14 \) or \( 6x + 10y \ge 1400 \) or \( 3x + 5y \ge 700 \). So, we need to minimise the cost \( Z = 6x + 5y \) with these constraints: \( 2x + y \ge 280 \), \( 3x + 5y \ge 700 \), \( x, y \ge 0 \). (i) The line \( 2x + y = 280 \) passes through \( A(140, 0) \) and \( B(0, 280) \). When \( x = 0, y = 0 \) is put into \( 2x + y \ge 280 \), we get \( 0 \ge 280 \), which is false.
\( \implies \) The region represented by \( 2x + y \ge 280 \) lies on and above AB. (ii) The line \( 3x + 5y = 700 \) passes through \( C(\frac{700}{3}, 0) \) and \( D(0, 140) \). When \( x = 0, y = 0 \) is put into \( 3x + 5y \ge 700 \), we get \( 0 \ge 700 \), which is false.
\( \implies \) The region represented by \( 3x + 5y \ge 700 \) lies on and above CD. (iii) \( x \ge 0 \) lies on and to the right of the y-axis. (iv) \( y \ge 0 \) lies on and above the x-axis. The shaded area YBPCX is the feasible region, where P is the point where AB and CD intersect. The lines are: \( 2x + y = 280 \) .......... (1) \( 3x + 5y = 700 \) .......... (2) Multiplying equation (1) by 5 and subtracting (2) from it, we get: \( 10x + 5y - (3x + 5y) = 1400 - 700 \) \( 7x = 700 \) \( \implies x = 100 \) From (1), \( 2(100) + y = 280 \) \( 200 + y = 280 \) \( y = 280 - 200 \) \( \implies y = 80 \) So, the point P is \( (100, 80) \). The values of Z at the corner points B, P, C are: At B\( (0, 280) \), \( Z = 6(0) + 5(280) = 0 + 1400 = 1400 \) At P\( (100, 80) \), \( Z = 6(100) + 5(80) = 600 + 400 = 1000 \) At C\( (\frac{700}{3}, 0) \), \( Z = 6(\frac{700}{3}) + 0 = 1400 \) So, Z is minimum at P\( (100, 80) \), where \( Z = 1000 \). The minimum cost of Z is Rs. 1000, when quantities of fertiliser F\( _1 \) and F\( _2 \) are 100 kg and 80 kg respectively.In simple words: To provide the right amount of nutrients at the lowest cost, a farmer should use 100 kg of fertiliser F1 and 80 kg of fertiliser F2. The lowest cost will be Rs. 1000.

Exam Tip: Always double-check your calculations for the intersection point, as an error here will lead to incorrect optimal values for the objective function.

 

Question 11. The corner points of the feasible region determined by the following system of linear inequalities: 2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5) Let Z = px + qy, where p, q > 0. Conditions on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is
(A) p = q
(B) p = 2q
(C) p = 3q
(D) q = 3p
Answer: (D) q = 3p
Maximum value of \( Z = px + qy \) occurs at both \( (3, 4) \) and \( (0, 5) \). At \( (3, 4) \), \( Z = p(3) + q(4) = 3p + 4q \). At \( (0, 5) \), \( Z = p(0) + q(5) = 5q \). Since both are maximum values, they must be equal: \( 3p + 4q = 5q \) \( 3p = 5q - 4q \) \( 3p = q \) Therefore, \( q = 3p \).In simple words: For the maximum profit to happen at two different points, the relationship between p and q must be that q is three times p.

Exam Tip: When the maximum (or minimum) value of an objective function occurs at two distinct corner points, it means the entire line segment joining those two points yields the same optimal value. This implies that the objective function line is parallel to that segment. To find the condition, simply set the Z-values at these two points equal to each other.