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Q.
$Z = 7x + y$, subject to $5x + y \ge 5, x + y \ge 3, x \ge 0, y \ge 0$. The minimum value of Z occurs at
Linear Programming
Solution:
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 $A\left(3, 0\right), \,B\left(\frac{1}{2}, \frac{5}{2}\right), C\left(0, 5\right)$ For B : 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 $5$
