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Q.
$Z = 6 x + 21 y$, subject to $x + 2 y \ge 3$, $x + 4 y\ge 4$, $3 x + y \ge 3$, $x \ge 0$, $y \ge 0$. The minimum value of $Z$ occurs at
Linear Programming
Solution:
We have, minimize $Z = 6x + 21y$
Subject to $x + 2y \ge 3, x + 4y \ge 4, 3x + y \ge 3, x \ge 0, y\ge 0$
Let $l_1 : x + 2y = 3, l_2 : x + 4y = 4, l_3 : 3x + y = 3$
$l_4 : x - 0$ and $l_5 : y = 0 $ For B : Solving $l_1$ and $l_2$, we get $B(2,1/2) $ For C : Solving $l_1$ and $l_3$, we get $C(0.6,1.2$)
Shaded portion is the feasible region, where $A(4, 0)$,
$B \left(2, \frac{1}{2}\right) C (0 .6 ,1 .2 ), D (0,3)$.
Now, minimize $Z = 6x + 21y$
$Z$ at $A(4, 0) = 6(4) + 21(0) = 24$
$Z$ at $B\left(2, \frac{1}{2}\right)=6(2) + 21(\frac {1}{2})=22.5 $
$Z$ at $C (0.6, 1.2) = 6(0.6) + 21(1.2) = 3.6 + 25.2 = 28.8$
$Z$ at $D(0, 3) = 6(0)+ 21(3) = 63 $
Thus, $Z$ is minimized at $\left(2, \frac{1}{2}\right)$ and its minimum value is $22.5$.
