Constraint for the Linear Programming (LP):

Constraint 1:

2x1+x2≥6

x₁ 0 3
x₂ 6 0

Constraint 2:

x1+x2≥4

x₁ 0 4
x₂ 4 0

Constraint 3:

2x1+10x2≥20

x₁ 0 10
x₂ 2 0

Graph for the above constraint:

The value of the objective function at each of these extreme points are as follows:

Extreme-coordinates Lines-through-extremes
Objective-function-value

Z=5x₁+x₂
A(0, 6) 2x1+x2≥</

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Chapter 3, Problem 25RP

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Solution Summary: The author explains that the minimum value of the objective function Z = 6 occurs at extreme point (0, 6) Therefore, the optimal solution to the given LP problem is:

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