Chapter 6 Linear Programming Miscellaneous Exercise 6 Solutions

(a) at the centre of feasible region

(c) If LPP has two optional solutions then it has infinitely many solutions

(b) a function to be maximized or minimized

(c)

(a) 56

(d) (40, 15)

(c) at vertex of feasible region

(b) All of the given function

(a) x = 0, y =

(b) (0, 0), (, 0), (3, 1), (0, 4)

(b) (2, 0)

(b) 13

(a) (2, 2)

(c) (0, 0)

(c) (3, 4)

First

First

End

Third and Fourth

y > x

x ≥ 3 and y ≥ 2

x ≥ 2y

x ≤ 40

True

False

False

True

True

True

True

OAED is the feasible region O(0, 0) A(0, 6) D (6, 0) and E is the intersection of 2x1 + 3x2 = 18 and 2x1 + x2 = 12 For E, Solving, 2x1 + 3x2 = 18 …….(i) 2x1 + x2 = 12 ……(ii) We get x1 = 4.5, x2 = 3 ∴ E = (4.5, 3) ∴ Maximum value of z = 40.5 at E(4.5, 3)

AED is the feasible region A(0, 27), D(21, 0) and E is the point of intersection of 3x + y = 27 and x + y = 21 For E, Solving 3x + y = 27 ………(i) x + y = 21 …….(ii) We get x = 3, y = 18 ∴ E(3, 18) ∴ Minimum value of z = 48 at (3, 18)

OAED is the feasible region; O(0, 0), A (0, 2) D (3, 0) and E is the point of intersection of 3x + 5y = 10 and 5x + 3y = 15 For E, Solving 3x + 5y = 10 5x + 3y = 15 We get, x = , y = ∴ E(, ) Since the maximum value of z = 20 at two points i.e, at A(0, 2) and E(, ). z is maximum at all points on segment AE. Hence it has infinite number of solutions.

Shaded portion CE is the feasible region Where C = (0, 3) and E is the point of intersection of x – y = 1 and x + y = 3 For E, Solving x – y = 1 …….(i) x + y = 3 ………(ii) We get, x = 2, y = 1 ∴ E(2, 1) ∴ Minimum value z = 7 at E(2, 1)

OCEB is the feasible region where O(0, 0), C(0, 6), B(5, 0) E is the point of intersection of 3x1 + x2 = 15 and 3x1 + 4x2 = 24 For E, Solving 3x1 + x2 = 15 ……..(i) 3x1 + 4x2 = 24 …….(ii) We get, x1 = 4, x2 = 3 ∴ E(4, 3) ∴ Maximum value of z = 25 at (4, 3)

OAED is the feasible region O(0, 0), A(0, 20), D(20, 0) E is x + 2y = 40 and 3x + 2y = 60 For E, Solving x + 2y = 40 3x + 2y = 60 We get, x = 10, y = 15 ∴ E = (10, 15) ∴ Maximum value of z = 1350 at E(10, 15).

AGHF is the feasible region where A(0, 27) F(30, 0) G is the point of intersection of 3x + y = 27 and x + y = 21 H is the point of intersection of x + y = 21 and x + 2y = 30 For G, Solving 3x + y = 27 …….(i) x + y = 21 ………(ii) We get, x = 3, y = 18 ∴ G(3, 8) For H, Solving x + y = 21 …….(i) x + 2y = 30 ……..(ii) We get, x = 12, y = 9 ∴ G = (12, 9) ∴ Maximum value of z = 48 at G(3, 18)

Let z be the profit which can be made by selling x chair and y table. ∴ x ≥ 0, y ≥ 0 Total profit = 140x + 210y According to the table, the constraints can be written as 3x + 3y ≤ 36 5x + 2y ≤ 50 2x + 6y ≤ 60 ∴ The given LPP can be formulated as. Maximize z = 140x + 210y Subject to 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0. OEGH is the feasible region O(0, 0), D(10.4, 0), E(0, 10) For G, Solving 3x + 3y = 36 ……..(i) 2x + 6y = 60 …….(ii) ∴ G = (3, 9) For H, Solving 5x + 2y = 52 ………(i) 3x + 3y = 36 ……..(ii) We get, x = , y = ∴ H(, ) ∴ z in maximum at (3, 9) and maximum profit = ₹ 2310.

Let x number of bicycle and y number of tricycle has to be manufactured and to be sold to get the profit (z) ∴ x ≥ 0, y ≥ 0 Total profit = 180x + 220y. The given LPP can be tabulated as follows: ∴ The given LPP can be formulated as Maximize z = 180x + 220y Subject to 6x + 4y ≤ 120, 3x + 10y ≤ 180, x ≥ 0, y ≥ 0 OCEB is the feasible region where O(0, 0) C(0, 18) B(20, 0) E is the point of intersection of 6x + 4y = 120 and 3x + 10y = 180 For E, Solving 6x + 4y = 120 ……..(i) 3x + 10y = 180 …….(ii) We get, x = 10, y = 15 ∴ E(10, 15) ∴ Maximum value of z is 5100 at E(10, 15) Hence 10 bicycles and 15 tricycles should be produced to get maximum profit.

Let x be the no. of units of chemicals, A produced and y be the no. of units of chemical B produced. Total cost is 4x + 6y The LPP is. Minimise z = 4x + 6y Subject to x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0. The shaded region BEC in the feasible region B(80, 0) C(0, 75) and E is the point of intersection of x + 2y = 80 and 3x + y = 75 For E, Solving x + 2y = 80 …….(i) 3x + y = 75 …….(ii) We get, x = 14, y = 33 ∴ E(14, 33) ∴ z is minimum at E(14, 33) and the minimum value of z = 254. Hence 14 units of chemical A and 33 units of chemical B are produced to get a minimum cost of ₹ 254.

Let x be the no. of mixers produced and y be the no.of food processors produced. The profit is 2000x + 3000y The LPP is Maximize 2 = 2000x + 3000y Subject to 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60 x, y ≥ 0. Let 3x + 3y = 36, i.e. x + y = 12 x = 0, y = 12, (0, 12) y = 0, x = 12, (12, 0) 5x + 2y = 50 x = 0, y = 25, (0, 25) y = 0, x = 10, (10, 0) Let 2x + 6y = 60, x + 3y = 30 x = 0, y = 10 (0, 10) y = 0, x = 30 (30, 0) The Shaded region OABCD is the feasible region. O(0, 0) A (10, 0) D(0, 10) B is the intersection of x + y = 12 and 5x + 2y = 50 Solving we get x = 8.67, y = 3.33 ∴ B(8.67, 3.33) C is the intersection of x + y = 12 and x + 3y = 30 Solving we get x = 3, y = 9 ∴ C(3, 9) ∴ Maximum value of z is 330000 at C(3, 9) Hence 3 mixers and 9 food processors should be produced to get a maximum profit of ₹ 33,000.

Let x be the no. of units of compound I used and y be the no of units of compound II used. The data can be tabulated as The LPP is minimize z = 800x + 640y Subject to 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18, x, y ≥ 0. Let 4x + 2 y = 16, 2x + y = 8 x = 0, y = 8, (0, 8) y = 0, x = 4, (4, 0) 12x + 2y = 24, 6x + y = 12 x = 0, y = 12, (0, 12) y = 0, x = 2, (2, 0) 2x + 6y = 18 x + 3y = 9 x = 0, y = 3, (0, 3) y = 0, x = 9, (9, 0) The Shaded region ABCD is the feasible region. A(0, 12) D(9, 0) B is the intersection of 6x + y = 12 and 2x + y = 8 Solving we get x = 1, y = 6 ∴ B(1, 6) C is the intersection of 2x + y = 8 and x + 3y = 9 Solving we get x = 3, y = 2 ∴ C(3, 2) ∴ The minimum value of z = 3680 at (3, 2) Hence 3 unit of compound I and 2 units of compound II should be used to get the minimum cost of ₹ 3680.

Let x be the no. of gift A produced and y be the no. of gift B produced. The data can be tabulated as The LLP is maximize z = 75x + 125y Subject to 4x + 2y ≤ 208, 2x + 4y ≤ 152, x, y ≥ 0 Let 4x + 2y = 208, 2x + 4y = 152, 2x + y = 104, x + 2y = 76, x = 0, y = 104; (0, 104), x = 0, y = 38; (0, 38) y = 0, x = 52; (52, 0), y = 0, x = 76; (76, 0) The shaded region OABC is the feasible region. O(0,0) A (52, 0) C(0, 38) B is the intersection of 2x + y = 104 and x + 2y = 76 Solving we get x = 44, y = 16 ∴ B(44, 16) ∴ Maximum value of z = 5300 at B(44, 16) Hence he should produce 44 gifts of type A & 16 of type B to get a maximum profit of ₹ 5300.

Let x denote the number of units of product A and y denote the number of units of product B. The total profit in ₹ 3x + 4y, The given statements can be tabulated as The constraints are x + y ≤ 450, 2x + y ≤ 600, x ≥ 0, y ≥ 0 ∴ The LPP can be formulated as Maximize z = 3x + 4y Subject to x + y ≤ 450, 2x + y ≤ 600, x ≥ 0, y ≥ 0. OAED in the feasible region O(0, 0) A(0, 450) D(300, 0) E is the intersection of x + y = 450 and 2x + y = 600 For E, Solving x + y = 450 ……..(i) 2x + y = 600 ……..(ii) Solving x = 150, y = 300 ∴ E(150, 300) ∴ Maximum value of z = 1800 at (0, 450)

Let x be the no. of electrical item A and y be the no. of electrical items to be manufactured per month to maximize the profit. Total profit is ₹ 20x + 30y The given condition can be tabulated as. The given LPP can be formulated as Maximize z = 20x + 30y Subject of 3x + 2y ≤ 210, 2x + 4y ≤ 300, x ≥ 0, y ≥ 0. BOCE is the feasible region B(70, 0) O(0, 0) C(0, 75) E is the intersection of 3x + 2y = 210 and 2x + 4y = 300 For E, Solving 3x + 2y = 210 ……….(i) 2x + 4y = 300 …….(ii) We get x = 30, y = 60 ∴ E(30, 60) ∴ Maximum value of z = ₹ 2400 at (30, 60) Hence 30 units of A and 60 units of B should be manufactured per month to get the maximum profit of ₹ 2400