Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 12 Linear Programming Miscellaneous Exercise Questions and Answers.
PSEB Solutions for Class 12 Maths Chapter 12 Miscellaneous Exercise
Question 1.
A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol.
How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
Solution.
Let the diet contain x and y packets of foods P and Q respectively.
Therefore, x ≥ 0 and y ≥ 0.
The mathematical formulation of the given problem is as follows.
Maximise Z = 6x + 3y ……………..(i)
subject to the constraints, 4x + y ≥ 80 ………… (ii)
x + 5y ≥ 115 …………..(iii)
3x + 2y ≤ 150 ……………(iv)
x, y ≥ 0 …………….(v)
The feasible region determined by the system of constraints is as follows.
The comer points of the feasible region are A (15, 20), B (40, 15) and C (2, 72).
The values of Z at these corner points are as follows.
Thus, the maximum value of Z is 285 at (40, 15).
Therefore to maximise the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used. The maximum amount of vitamin A in the diet is 285 units.
Question 2.
A farmer mixes two brands P and Q of cattle feed. Brand P costing ₹ 250 per bag contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing ₹ 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B and 3 units of element C.
The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Solution.
Let the farmer mix x bags of brand P and y bags of brand Q.
The given information can be compiled in a table as follows
The given problem can be formulated as follows.
Minimise Z = 250x + 200y …………(i)
subject to the constraints, 3x + 1.5y ≥ 18 …………(ii)
2.5x + 11.25y ≥ 45 …………(iii)
2x + 3y ≥ 24 …………..(iv)
x, y ≥ 0 ………….(v)
The feasible region determined by the system of constraints is as follows.
The comer points of the feasible region are A (18, 0) B (9, 2) C (3, 6) and D (0, 12).
The values of Z at these corner points are as follows.
As the feasible region is unbounded, therefore, 1950 may or may not be minimum value of Z.
For this, we draw a graph of the inequality, 250x + 200y < 1950 or 5x + 4y < 39 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 5x + 4y < 39.
Therefore, the minimum value of Z is 2000 at (3, 6).
Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost to ₹ 1950.
Question 3.
A dietician wishes to mix together 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 ₹ 16 and one kg of food Y costs ₹ 20. Find the least cost of the mixture which will produce the required diet?
Solution.
Let the mixture contain x kg of food X and y kg of food Y.
The mathematical formulation of the given problem is as follows,
Minimise Z = 16x + 20y ………….(i)
subject to the constraints, x + 2y ≥ 10 ……… (ii)
x + y ≥ 6 …………….(iii)
3x + y ≥ 8 ………….(iv)
x, y ≥ 0 ………….(v)
The feasible region determined by the system of constraints is as follows.
The corner points of the feasible region are A (10, 0) B (2, 4), C (1, 5) and D (0, 8).
The values of Z at these corner points are as follows.
As the feasible region is unbounded, therefore, 112 may or may not be the minimum value of Z.
For this, we draw a graph of the.. inequality, 16x + 20y <112 or 4x + 5y < 28 and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 4x + 5y < 28.
Therefore, the minimum value of Z is 112 at (2, 4).
Thus, the mixture should contain 2 kg of food X and 4 kg of food Y.
The mixture is ₹ 112.
Question 4.
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below.
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is ₹ 7.50 and that on each toy of type B is ₹ 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Solution.
Let x and y toys of type A and type B respectively be manufactured in a day.
The given problem can be formulated as follows.
Maximise Z = 7.5x + 5y ……………….(i)
subject to the constraints, 2x + y ≤ 60 …………….(ii)
x ≤ 20 ………………..(iii)
2x + 3y ≤ 120 ……………..(iv)
x, y ≥ 0 …………………(v)
The feasible region determined by the constraints is as follows.
The corner points of the feasible region are A (20, 0), B (20, 20) C (15, 30) and D (0, 40).
The values of Z at these comer points are as follows.
The values of Z is 262.5 at (15, 30).
Thus, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximise the profit.
Question 5.
An aeroplane can carry a maximum of 200 passengers. A profit of ₹ 1000 is made on each executive class ticket and a profit of ₹ 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Solution.
Let the airline sell x tickets of executive class and y tickets of economy class.
The mathematical formulation of the given problem is as follows.
Maximise Z = 1000x + 600y ……………..(i)
subject to the constraints, x + y ≤ 200 …………(ii)
x ≥ 20 …………(iii)
y – 4x ≥ 0 ……….(iv)
x, y ≥ 0 ……………..(v)
The feasible region determined by the constraints is as follows.
The corner points of the feasible region are A (20, 80), B (40, 160) and C (20, 180).
The values of Z at these corner points are as follows.
The maximum value of Z is 136000 at (40, 160).
Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximise the profit and the maximum profit is ₹ 136000.
Question 6.
Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table.
How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost? Solution.
Let godown A supply x and y quintals of grain to the shops D and E respectively. Then, (100 – x – y) will be supplied to shop F.
The requirement at shop D is 60 quintals since x quintals are transported from godown A.
Therefore, the remaining (60 – x) quintals will be transported from godown. B.
Similarly, (50 – y) quintals and 40 – (100 – x – y) = (x + y – 60) quintals will be transported from godown B to shop E and F respectively.
The given problems can be represented diagrammatically in above figure x ≥ 0, y ≥ 0 and 100 – x – y ≥ 0
=» x ≥ 0, y ≥ 0 and x + y ≤ 100 60 – x ≥ 0, 50 – y ≥ 0 and x + y – 60 ≥ 0
⇒ x ≤ 60, y ≤ 50 and x + y ≥ 60
Total transportation cost Z is given by,
Z = 6x + 3y + 2.5 (100 – x – y) + 4 (60 – x) + 2 (50 – y) + 3 (x + y – 60)
= 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x +100 – 2y + 3x + 3y -180
= 2.5x + 1.5y + 410
The given problem can be formulated as
Minimise Z = 2.5x + 1.5y + 410 …………..(i)
subject to the constraints, x + y ≤ 100 ………..(ii)
x ≤ 60 …………..(iii)
y ≤ 50 …………(iv)
x + y ≥ 60 ………….(v)
x, y ≥ 0 ……………(vi)
The feasible region determined by the constraints is as follows.
The values of Z at these corner points are as follows.
The minimum value of Z is 510 at (10, 50).
Thus, the amount of grain transported from A to D, E and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E and F is 50 quintals, 0 quintals, and 0 quintals respectively.
The minimum cost is ₹ 510.
Question 7.
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 and F whose requirements are 4500 L, 3000 L and 3500 L respectively. The distance (in km) between the depots and the petrol pumps is given in the following table:
Assuming that the transportation cost of 10 litres of oil is ₹ 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
Solution.
Let x and y litres of oil be supplied from A to the petrol pumps, D and E. Then, (7000 – x – y) will be supplied from A to petrol pump F.
The requirement at petrol pump D is 4500 L.
Since xL are transported from depot A, the remaining (4500 – x) L will be transported from petrol pump B.
Similarly, (3000 – y) L and 3500 – (7000 – x – y) = (x + y -3 500) L will be transported from depot B to petrol E and F respectively.
The given problems can be represented diagrammatically as follows.
x ≥ 0, y ≥ 0 and (7000 – x – y) ≥ 0
⇒ x ≥ 0, y ≥ 0 and x + y ≤ 7000
4500 – x ≥ 0, 3000 – y ≥ 0 and x + y – 3500 ≥ 0
⇒ x ≤ 4500, y ≤ 3000 and x + y ≥ 3500
Cost of transporting 10 L of petrol = ₹ 1
Cost of transporting 1 L of petrol = ₹ \(\frac{1}{10}\)
Therefore total transportation cost is given by,
Z = \(\frac{7}{10}\) x + \(\frac{6}{10}\) y + \(\frac{3}{10}\) (7000 – x – y) + \(\frac{3}{10}\) (4500 – x) + \(\frac{4}{10}\) (3000 – y) + \(\frac{2}{10}\) (x + y – 3500)
= 0.3x + 0.1 y + 3950
The problem can be formulated as follows
Minimise Z = 0.3x + 0.1y + 3950 …………….(i)
subjectto the constraints, x + y ≤ 7000 ………….(ii)
x ≤ 4500 ……………(iii)
y ≤ 3000 ………….(iv)
x + y ≥ 350 …………….(v)
x, y ≥ 0 ………….(vi)
The feasible region determined by the constraints is as follows.
The corner points of the feasible region are A (3500, 0), B (4500, 0) C (4500, 2500), D (1000, 3000) and E (500, 3000).
The values of Z at these corner points are as follows.
The minimum value of Z is 4400 at (500, 3000).
Thus, the oil supplied from depot A is 500 L, 3000 L and 3500 L, and from depot B is 4000 L, 0L and 0 L to petrol pumps D, E and F respectively. The minimum transportation cost is ₹ 4400.
Question 8.
A fruit grower can use two types of fertiliser in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid atleast 270 kg of potash and at most 310 kg of chlorine.
If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
Solution.
Let the fruit grower use x bags of brand P and y bags of brand Q.
The problem can be formulated as follows.
Minimise Z = 3x + 3.5y
subject to the constraints, x + 2y ≥ 240 ………….(ii)
x + 0.5y ≥ 90 ……………….(iii)
1.5x + 2y ≤ 310 …………….(iv)
x, y ≥ 0
The feasible region determined by the system of constraints is as follows.
The corner points are A(140, 50), B(20, 140) and C(40, 100).
The values of Z at these corner points are as follows.
The minimum value of Z is 470 at (40, 100).
Thus, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimise the amount of nitrogen.
The minimum amount of nitrogen added to the garden is 470 kg.
Question 9.
Refer to question 8. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
Solution.
Let the fruit grower use x bags of brand P and y bags of brand Q.
The problem can be formulated as follows.
Maximise Z = 3x + 3.5y ……………(i)
subject to the constraints, x + 2y ≥ 240 …………..(ii)
x + 0.5y ≥ 90 …………(iii)
1.5x + 2y ≤ 310 …………..(iv)
x, y ≥ 0 ……….(v)
The feasile region determined y the system of constraints is as follows.
The corner points are A (1. 40, 50), B(20, 140) and C(40, 100).
The values of Z at these corner points are as follows.
The maximum value of Z is 595 at (140, 50).
Thus, 140 bags of brand P and 50 bags of brand Q should be used to maximise the amount of nitrogen.
The maximum amount of nitrogen added to the garden is 595 kg.
Question 10.
A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A.
Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of ₹ 12 and 116 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
Solution.
Let x and y be the number of dolls of type A and B respectively that are produced per week.
The given problem can be formulated as follows.
Maximise Z = 12x + 16y ……………(i)
subject to the constraints, x + y ≤ 1200……….(ii)
y ≤ \(\frac{x}{2}\)
⇒x ≥ 2y ………..(iii)
x – 3y ≤ 600 ……………(iv)
x, y ≥ 0 …………….(v)
The feasible region determined by the system of constraints is as follows.
The corner points are A(600, 0), B(1050, 150) and C(800, 400).
The Y’ values of Z at these corner points are as follows.
The maximum value of Z is 16000 at (800, 400).
Thus, 800 and 400 dolls of type A and type B should be produced respectively to get the maximum profit of ₹ 16000.