Maximize z = 11 x + 8 y , subject to x ≤ 4 , y ≤ 6 , x + y ≤ 6 , x ≥ 0 , y ≥ 0 .

Inequalities x ≤ 4 y ≤ 6 x + y ≤ 6 Equalities x = 4 y = 6 x + y = 6 Intercept form x 4 = 1 y 6 = 1 x 6 + y 6 = 1 Origin Test 0 ≤ 4 0 ≤ 6 0 + 0 ≤ 6 True Origin Side True Origin Side True Origin Side Shaded portion O A B C is the feasible region, where O ( 0 , 0 ) , A ( 4 , 0 ) , C ( 0 , 6 ) For B : put x = 4 in x + y = 6 ∴ 4 + y = 6 ⇒ y = 2 Thus, B ( 4 , 2 ) z = 11 x + 8 y z ( O ) = ( 0 , 0 ) = 11 ( 0 ) + 8 ( 0 ) = 0 z ( A ) = ( 4 , 0 ) = 11 ( 4 ) + 8 ( 0 ) = 44 z ( B ) = ( 4 , 2 ) = 11 ( 4 ) + 8 ( 2 ) = 60 z ( C ) = ( 0 , 6 ) = 11 ( 0 ) + 8 ( 6 ) = 48 Thus, z is maximized at B ( 4 , 2 ) and its maximum value is 60 .

Solve the following L.P.P. by graphical method: Maximize: z = 10 x + 25 y subject to 0 ≤ x ≤ 3 , 0 ≤ y ≤ 3 , x + y ≤ 5 also find maximum value of z .

Given, To maximize: z = 10 x + 25 y subject to 0 ≤ x ≤ 3 , 0 ≤ y ≤ 3 , x + y ≤ 5 , Now solving the equations in table form we get, Inequalities x ≤ 3 y ≤ 3 x + y ≤ 5 Equations x = 3 y = 3 x + y = 5 Origin Test 0 ≤ 3 0 ≤ 3 0 + 0 ≤ 5 True Origin Side True Origin Side True Origin Side Now plotting the diagram and sketching feasible region we get, Shaded portion O A B C D is the feasible region, where O ( 0 , 0 ) , A ( 3 , 0 ) , D ( 0 , 3 ) , B ( 3 , 2 ) , C ( 2 , 3 ) z = 10 x + 25 y z ( O ) = ( 0 , 0 ) = 10 ( 0 ) + 25 ( 0 ) = 0 z ( A ) = ( 3 , 0 ) = 10 ( 3 ) + 25 ( 0 ) = 30 z ( B ) = ( 3 , 2 ) = 10 ( 3 ) + 25 ( 2 ) = 80 z ( C ) = ( 2 , 3 ) = 10 ( 2 ) + 25 ( 3 ) = 95 z ( D ) = ( 0 , 3 ) = 10 ( 0 ) + 25 ( 3 ) = 75 Thus, z is maximized at C ( 2 , 3 ) and its maximum value is 95 .

Maximize z = 3 x + 5 y , subject to x + 4 y ≤ 24 , 3 x + y ≤ 21 , x + y ≤ 9 , x ≥ 0 , y ≥ 0 Also find maximum value of z .

Inequalities x + 4 y ≤ 24 3 x + y ≤ 21 x + y ≤ 9 Equalities x + 4 y = 24 3 x + y = 21 x + y = 9 Intercept form x 24 + y 6 = 1 x 7 + y 21 = 1 x 9 + y 9 = 1 Origin Test 0 + 4 ( 0 ) ≤ 24 0 ≤ 6 0 + 0 ≤ 9 True Origin Side True Origin Side True Origin Side Shaded portion O A B C D is the feasible region, where O ( 0 , 0 ) , A ( 7 , 0 ) , D ( 0 , 6 ) For B : 3 x + y = 21 x + y = 9 ⇒ x = 9 - y ∴ 3 9 - y + y = 21 ⇒ 27 - 3 y + y = 21 ⇒ - 2 y = - 6 ⇒ y = 3 ∴ x = 9 - 3 = 6 ∴ B ( 6 , 3 ) For C : x + 4 y = 24 x + y = 9 ⇒ x = 9 - y ∴ 9 - y + 4 y = 24 ⇒ 3 y = 15 ⇒ y = 5 ∴ x = 9 - 5 = 4 ∴ C = ( 4 , 5 ) z = 3 x + 5 y z ( O ) = ( 0 , 0 ) = 3 ( 0 ) + 5 ( 0 ) = 0 z ( A ) = ( 7 , 0 ) = 3 ( 7 ) + 5 ( 0 ) = 21 z ( B ) = ( 6 , 3 ) = 3 ( 6 ) + 5 ( 3 ) = 33 z ( C ) = ( 4 , 5 ) = 3 ( 4 ) + 5 ( 5 ) = 37 z ( D ) = ( 0 , 6 ) = 3 ( 0 ) + 5 ( 6 ) = 30 Thus, z is maximized at C ( 4 , 5 ) and its maximum value is 37 .

E M B I B E

Mathematics & Statistics (Arts & Science) Part 1 Standard 12>Chapter 7 - Linear Programming>Exercise 7.4

Maharashtra Board Solutions for Chapter: Linear Programming, Exercise 4: Exercise 7.4

Author:Maharashtra Board

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Linear Programming, Exercise 4: Exercise 7.4

Attempt the practice questions on Chapter 7: Linear Programming, Exercise 4: Exercise 7.4 with hints and solutions to strengthen your understanding. Mathematics & Statistics (Arts & Science) Part 1 Standard 12 solutions are prepared by Experienced Embibe Experts.