# Solve the LPP graphically:Minimize Z = 4x + 5ySubject to the constraints 5x + y ≥ 10, x + y ≥ 6, x + 4y ≥ 12, x, y ≥ 0 Solution: Convert the constraints into equations and find the intercept made by - Mathematics and Statistics

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Solve the LPP graphically:
Minimize Z = 4x + 5y
Subject to the constraints 5x + y ≥ 10, x + y ≥ 6, x + 4y ≥ 12, x, y ≥ 0

Solution: Convert the constraints into equations and find the intercept made by each one of it.

 Inequations Equations X intercept Y intercept Region 5x + y ≥ 10 5x + y = 10 ( ___, 0) (0, 10) Away from origin x + y ≥ 6 x + y = 6 (6, 0) (0, ___ ) Away from origin x + 4y ≥ 12 x + 4y = 12 (12, 0) (0, 3) Away from origin x, y ≥ 0 x = 0, y = 0 x = 0 y = 0 1st quadrant

∵ Origin has not satisfied the inequations.

∴ Solution of the inequations is away from origin.

The feasible region is unbounded area which is satisfied by all constraints.

In the figure, ABCD represents

The set of the feasible solution where

A(12, 0), B( ___, ___ ), C ( ___, ___ ) and D(0, 10).

The coordinates of B are obtained by solving equations

x + 4y = 12 and x + y = 6

The coordinates of C are obtained by solving equations

5x + y = 10 and x + y = 6

Hence the optimum solution lies at the extreme points.

The optimal solution is in the following table:

 Point Coordinates Z = 4x + 5y Values Remark A (12, 0) 4(12) + 5(0) 48 B ( ___, ___ ) 4( ___) + 5(___ ) ______ ______ C ( ___, ___ ) 4( ___) + 5(___ ) ______ D (0, 10) 4(0) + 5(10) 50

∴ Z is minimum at ___ ( ___, ___ ) with the value ___

#### Solution

Convert the constraints into equations and find the intercept made by each one of it.

 Inequations Equations X intercept Y intercept Region 5x + y ≥ 10 5x + y = 10 (2, 0) (0, 10) Away from origin x + y ≥ 6 x + y = 6 (6, 0) (0, 6) Away from origin x + 4y ≥ 12 x + 4y = 12 (12, 0) (0, 3) Away from origin x, y ≥ 0 x = 0, y = 0 x = 0 y = 0 1st quadrant

∵ Origin has not satisfied the inequations.

∴ Solution of the inequations is away from origin.

The feasible region is unbounded area which is satisfied by all constraints.

In the figure, ABCD represents

The set of the feasible solution where

A(12, 0), B(4, 2), C (1, 5) and D(0, 10).

The coordinates of B are obtained by solving equations

x + 4y = 12 and x + y = 6

The coordinates of C are obtained by solving equations

5x + y = 10 and x + y = 6

Hence the optimum solution lies at the extreme points.

The optimal solution is in the following table:

 Point Coordinates Z = 4x + 5y Values Remark A (12, 0) 4(12) + 5(0) 48 B (4, 2) 4(4) + 5(2) 26 minimum C (1, 5) 4(1) + 5(5) 29 D (0, 10) 4(0) + 5(10) 50

∴ Z is minimum at B (4, 2) with the value 26

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Chapter 2.6: Linear Programming - Q.5 (E)
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