Question

$Z = 7x + y$, subject to $5x + y \ge 5, x + y \ge 3, x \ge 0, y \ge 0$. The minimum value of Z occurs at

• A. $(3, 0)$
• B. $\left(\frac{1}{2}, \frac{5}{2}\right)$
• C. $(7, 0)$
• D. $(0, 5)$

Answer 1

The Correct Option is D

We have, maximize $Z = 7x + y$, Subject to : $5x + y \ge 5, x + y \ge 3, x, y \ge 0$. Let $\ell_{1} : 5x + y = 5$ $\ell _{2} : x + y = 3$ $\ell _{3} : x = 0$ and $\ell _{4} : y = 0$ Shaded portion is the feasible region, Where $A\left(3, 0\right), \,B\left(\frac{1}{2}, \frac{5}{2}\right), C\left(0, 5\right)$

Solving $\ell _{1}$ and $\ell _{2}$, we get $B\left(\frac{1}{2}, \frac{5}{2}\right)$ Now maximize $Z = 7x + y$ Z at $A\left(3, 0\right) = 7\left(3\right) + 0 = 21$ Z at $B\left(\frac{1}{2}, \frac{5}{2}\right) = 7\left(\frac{1}{2}\right) + \frac{5}{2} = 6$ Z at $C\left(0, 5\right) = 7\left(0\right) + 5 = 5$ Thus Z, is minimized at $C\left(0, 5\right)$ and its minimum value is $5$