Question

The problem of maximizing $Z=x_1-x_2$ subject to constraints

$x_1+x_2\le 10,x_1\ge 0,x_2\ge 0\text{and}{x}_2\le 5\text{has}$

• A. two solutions
• B. no solution
• C. one solution
• D. more than two solutions

Answer 1

The correct option is C one solution


$Max.,Z=x_1-x_2$

$\text{Constrains:}$

$x_1+x_2\le 10;$

$x_1\ge 0;x_2\ge 0\text{and}{x}_2\le 5$

$\text{Z(0, 5) = 0 - 5 = -5}$

$\text{Z(5, 5) = 5 - 5 = 0}$

$\text{Z(10, 0) = 10 - 0 = 10}$

$Z_{max}=10at(10,0)$

$\therefore \text{The problem has one solution}$