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Q.
$Z = 8x + 10y$, subject to $2 x + y \ge 7$, $2 x + 3 y \ge 15$, $y \ge 2$, $x \ge 0$, $y \ge 0$. The minimum value of $Z$ occurs at
Linear Programming
Solution:
We have, $Z = 8x + 10y$,
subject to $2 x + y \ge 7, 2 x + 3 y \ge 15, y \ge 2, x \ge 0, y \ge 0$.
Let $l_1 : 2x + y = 7, l_2: 2x + 3 y = 15, l_3 : y = 2$
$ l_4 : x = 0, l_5 : y = 0$ For A: Solving $l_2$ and $l_3$, we get $A(4.5,2)$ For B : Solving $l_1$ and $l_2$, we get $B(1.5,4)$
Shaded portion is the feasible region, where
$A(4.5, 2), B(1.5,4),C(0,7)$
Now, minimize $Z = 8x+10y$
$Z$ at $A(4.5,2) = 8(4.5) + 10(2) = 56$
$Z$ at $B (1.5,4) = 8(1.5) + 10(4) = 52$
$Z$ at $C(0, 7) = 8 (0) + 10(7) = 70$
Thus, $Z$ is minimized at $B(1.5,4)$ and its minimum value is $52$.
