CBSE Class 12 Mathematics Linear Programming Important Questions Set 01

Objective Type Questions:

Question. The solution set of the inequation \( 3x + 2y > 3 \) is
(a) half plane not containing the origin
(b) half plane containing the origin
(c) the point being on the line \( 3x + 2y = 3 \)
(d) None of these
Answer: (a)

Question. If the constraints in a linear programming problem are changed
(a) solution is not defined
(b) the objective function has to be modified
(c) the problems is to be re-evaluated
(d) none of these
Answer: (c)

Question. Which of the following statement is correct?
(a) Every LPP admits an optimal solution.
(b) Every LPP admits unique optimal solution.
(c) If a LPP gives two optimal solutions it has infinite number of solutions.
(d) None of these
Answer: (c)

Question. The maximum value of \( p = x + 3y \) such that \( 2x + y \leq 20 \), \( x + 2y \leq 20 \), \( x \geq 0, y \geq 0 \) is
(a) 10
(b) 30
(c) 60
(d) \( \frac{80}{3} \)
Answer: (b)

Fill in the blanks.

Question. In a LPP, the linear function which has to be maximised or minimised is called a linear ____________ function.
Answer: objective

Question. The maximum value of \( Z = 6x + 16y \) satisfying the conditions \( x + y \geq 2 \), \( x \geq 0, y \geq 0 \) is ____________.
Answer: does not exist (unbounded region)

Question. In a LPP, the inequalities or restrictions on the variables are called ____________.
Answer: constraints

Very Short Answer Questions:

Question. Determine the maximum value of \( Z = 3x + 4y \), Subject to the constraints: \( x + y \leq 1, x \geq 0, y \geq 0 \).
Answer: 4

Question. If a linear programming problem is \( Z_{\text{max}} = 3x + 2y \), Subject to the constraints: \( x + y \leq 2 \), find \( Z_{\text{max}} \).
Answer: 6

Question. Maximise the function \( Z = 11x + 7y \), subject to the constraints: \( x \leq 3, y \leq 2, x \geq 0, y \geq 0 \).
Answer: 47

Short Answer Questions–I:

Question. A manufacturing company makes two models A and B of a product. Each piece of model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing the maximum labour hours available are 180 and 30 respectively. The company makes a profit of ₹ 8,000 on each piece of model A and ₹ 12,000 on each piece of model B. To realise a maximum profit formulate above problem in LPP.
Answer: Let \( x \) and \( y \) be the number of pieces of model A and model B respectively.
Maximise \( Z = 8000x + 12000y \)
Subject to constraints:
\( 9x + 12y \leq 180 \)
\( x + 3y \leq 30 \)
\( x \geq 0, y \geq 0 \)

Question. 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. To get the maximum number of cakes can be made from 5 kg of flour and 1 kg of fat, formulate the problem in LPP.
Answer: Let \( x \) be the number of cakes of the first kind and \( y \) be the number of cakes of the second kind.
Maximise \( Z = x + y \)
Subject to constraints:
\( 200x + 100y \leq 5000 \Rightarrow 2x + y \leq 50 \)
\( 25x + 50y \leq 1000 \Rightarrow x + 2y \leq 40 \)
\( x \geq 0, y \geq 0 \)

Question. 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 ₹ 17.50 per package on nuts and ₹ 7.00 per package on bolts. If he operates his machines for at most 12 hours a day, the formulate the problems in LPP to maximise his profit.
Answer: Let \( x \) be the number of packages of nuts and \( y \) be the number of packages of bolts.
Maximise \( Z = 17.50x + 7y \)
Subject to constraints:
\( x + 3y \leq 12 \)
\( 3x + y \leq 12 \)
\( x \geq 0, y \geq 0 \)

Question. A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost ₹ 25,000 and ₹ 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. If he does not want to invest more than ₹ 70 lakhs and if his profit on the desktop model is ₹ 4500 and on portable mode is ₹ 5000, then formulate the problems as LPP to get maximum profit.
Answer: Let \( x \) be the number of desktop models and \( y \) be the number of portable models.
Maximise \( Z = 4500x + 5000y \)
Subject to constraints:
\( x + y \leq 250 \)
\( 25000x + 40000y \leq 7000000 \Rightarrow 5x + 8y \leq 1400 \)
\( x \geq 0, y \geq 0 \)

Question. A manufacturing company makes two toys A and B. Each piece of toy A requires 8 labour hours for fabricating and 2 labour hours for finishing. Each piece of toy B requires 16 labour hours for fabricating and 4 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 200 and 50 respectively. The company makes a profit of ₹ 10,000 on each piece of toy A and ₹ 14,000 on each piece of toy B. Formulate the problem as LPP to realise a maximum profit.
Answer: Let \( x \) and \( y \) be the number of toys A and B respectively.
Maximise \( Z = 10000x + 14000y \)
Subject to constraints:
\( 8x + 16y \leq 200 \Rightarrow x + 2y \leq 25 \)
\( 2x + 4y \leq 50 \Rightarrow x + 2y \leq 25 \)
\( x \geq 0, y \geq 0 \)

Short Answer Questions–II:

Question. A cottage manufactures pedestal lamps and wooden shades. Both the products require machine time as well as craftsman time in the making. The number of hour(s) required for producing 1 unit of each and the corresponding profit is given in the following table:

In day, the factory has availability of not more than 42 hours of machine time and 24 hours of craftsman time.
Assuming that all items manufactured are sold, how should the manufacturer schedule his daily production in order to maximise the profit? Formulate it as an LPP and solve it graphically.
Answer: Let \( x \) pedestal lamps and \( y \) wooden shades be manufactured.
Maximise \( Z = 30x + 20y \)
Subject to:
\( 1.5x + 3y \leq 42 \Rightarrow x + 2y \leq 28 \)
\( 3x + y \leq 24 \)
\( x, y \geq 0 \)
Graphically, corner points are (0, 0), (8, 0), (0, 14), and (4, 12).
\( Z(0,0) = 0 \)
\( Z(8,0) = 240 \)
\( Z(0,14) = 280 \)
\( Z(4,12) = 120 + 240 = 360 \).
Max profit is ₹ 360 by producing 4 lamps and 12 shades.

Question. A manufacturer has three machines I, II and III installed in his factory. Machine I and II are capable of being operated for atmost 12 hours whereas machine III must be operated for atleast 5 hours a day. He produces only two items M and N each requiring the use of all the three machines.
The number of hours required for producing 1 unit of M and N on three machines are given in the following table:

He makes a profit of ₹ 600 and ₹ 400 on one unit of items M and N respectively. How many units of each item should he produce so as to maximise his profit assuming that he can sell all the items that he produced. What will be the maximum profit?
Answer: Let \( x \) units of M and \( y \) units of N be produced.
Maximise \( Z = 600x + 400y \)
Subject to:
\( x + 2y \leq 12 \)
\( 2x + y \leq 12 \)
\( x + 1.25y \geq 5 \)
\( x, y \geq 0 \)
Corner points are (0, 6), (4, 4), (6, 0), (5, 0), (0, 4).
\( Z(0,6) = 2400 \)
\( Z(4,4) = 2400 + 1600 = 4000 \)
\( Z(6,0) = 3600 \)
\( Z(5,0) = 3000 \)
\( Z(0,4) = 1600 \)
Maximum profit is ₹ 4000 by producing 4 units of M and 4 units of N.

Question. A man rides his motorcycle at the speed of 50 km/hour. He has to spend ₹ 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to ₹ 3 per km. He has at the most ₹ 120 to spend on petrol and one hour time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.
Answer: Let \( x \) km be the distance travelled at 50 km/h and \( y \) km at 80 km/h.
Maximise \( Z = x + y \)
Subject to:
\( 2x + 3y \leq 120 \) (Petrol constraint)
\( \frac{x}{50} + \frac{y}{80} \leq 1 \) (Time constraint)
\( x, y \geq 0 \)

Question. Solve the following linear programming problem graphically:
Minimise \( Z = x - 5y + 20 \)
Subject to constraints: \( x - y \geq 0 \), \( -x + 2y \geq 2 \), \( x \geq 3 \), \( y \leq 4 \), \( x, y \geq 0 \)
Answer: Graphical solution yields corner points (3, 2.5), (3, 3), (4, 4), (3, 4).
\( Z(3, 2.5) = 3 - 12.5 + 20 = 10.5 \)
\( Z(3, 3) = 3 - 15 + 20 = 8 \)
\( Z(4, 4) = 4 - 20 + 20 = 4 \)
\( Z(3, 4) = 3 - 20 + 20 = 3 \).
Minimum value is 3 at (3, 4).

Question. Solve the following LPP:
Maximise \( Z = 5x_1 + 7x_2 \)
Subject to constraints: \( x_1 + x_2 \leq 4 \), \( 3x_1 + 8x_2 \leq 24 \), \( 10x_1 + 7x_2 \leq 35 \), \( x_1, x_2 \geq 0 \).
Answer: Graph corners are (0,0), (3.5, 0), (1.6, 2.4), (0, 3).
\( Z(0,0) = 0 \)
\( Z(3.5, 0) = 17.5 \)
\( Z(1.6, 2.4) = 8 + 16.8 = 24.8 \)
\( Z(0, 3) = 21 \).
Maximum value is 24.8 at \( x_1 = 1.6, x_2 = 2.4 \).

Question. Maximise \( Z = x + y \) subject to \( x + y \leq 8 \), \( x + y \leq 4 \), \( 2x + 3y \leq 12 \), \( 3x + y \leq 9 \), \( x, y \geq 0 \).
Answer: Common region corner points are (0,0), (3,0), (2.1, 2.6), (0, 4).
\( Z(0,0) = 0 \)
\( Z(3,0) = 3 \)
\( Z(2.1, 2.6) = 4.7 \)
\( Z(0,4) = 4 \).
Maximum value is 4.7 at (2.1, 2.6).

Question. Solve the following LPP graphically:
Maximise \( Z = 1000x + 600y \)
Subject to the constraints \( x + y \leq 200 \), \( x \geq 20 \), \( y - 4x \geq 0 \), \( x, y \geq 0 \)
Answer: Corner points are (20, 80), (20, 180), (40, 160).
\( Z(20, 80) = 20000 + 48000 = 68000 \)
\( Z(20, 180) = 20000 + 108000 = 128000 \)
\( Z(40, 160) = 40000 + 96000 = 136000 \).
Maximum value is 1,36,000 at (40, 160).

Question. Solve the following LPP graphically:
Maximise \( Z = 4x + y \)
Subject to following constraints \( x + y \leq 50 \), \( 3x + y \leq 90 \), \( x \geq 10 \), \( x, y \geq 0 \)
Answer: Corner points are (10, 0), (30, 0), (20, 30), (10, 40).
\( Z(10,0) = 40 \)
\( Z(30,0) = 120 \)
\( Z(20,30) = 80 + 30 = 110 \)
\( Z(10,40) = 40 + 40 = 80 \).
Maximum value is 120 at (30, 0).

Question. Solve the following linear programming problem graphically:
Maximise \( Z = 7x + 10y \)
Subject to constraints \( 4x + 6y \leq 240 \), \( 6x + 3y \leq 240 \), \( x \geq 10 \), \( x, y \geq 0 \)
Answer: Corner points are (10, 0), (40, 0), (30, 20), (10, 100/3).
\( Z(10,0) = 70 \)
\( Z(40,0) = 280 \)
\( Z(30,20) = 210 + 200 = 410 \)
\( Z(10, 100/3) = 70 + 1000/3 \approx 403.33 \).
Maximum value is 410 at (30, 20).

Question. Find graphically, the maximum value of \( Z = 2x + 5y \), subject to constraints given below:
\( 2x + 4y \leq 8 \), \( 3x + y \leq 6 \), \( x + y \leq 4 \), \( x \geq 0, y \geq 0 \)
Answer: Corner points are (0,0), (2,0), (1.6, 1.2), (0, 2).
\( Z(0,0) = 0 \)
\( Z(2,0) = 4 \)
\( Z(1.6, 1.2) = 3.2 + 6 = 9.2 \)
\( Z(0,2) = 10 \).
Maximum value is 10 at (0, 2).

Question. Solve the following linear programming problem graphically.
Minimise \( Z = 3x + 5y \)
Subject to the constraints: \( x + 2y \geq 10 \); \( x + y \geq 6 \); \( 3x + y \geq 8 \); \( x, y \geq 0 \)
Answer: Region is unbounded. Corner points are (0, 8), (1, 5), (2, 4), (10, 0).
\( Z(0,8) = 40 \)
\( Z(1,5) = 3 + 25 = 28 \)
\( Z(2,4) = 6 + 20 = 26 \)
\( Z(10,0) = 30 \).
Minimum value is 26 at (2, 4).