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  4. Z = 4x1 + 5x2, subject to 2x1 + x2 ge 7, 2x1 + 3x2 le 15, x2 le 3, x1, x2 ge 0. The minimum value of Z occurs at

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Q. $Z = 4x_1 + 5x_2$, subject to $2x_1 + x_2 \ge 7$, $2x_1 + 3x_2 \le 15$, $x_2 \le 3$, $x_1$, $x_2 \ge 0$. The minimum value of $Z$ occurs at

Linear Programming

Solution:

We have, minimize $Z = 4x_1 + 5x_2$
subject to $2x_1 + x_2 \ge 7, 2x_1 + 3x_2 \le 15, x_2 \le 3, x_1, x_2 \ge 0$
Let $l_1$ : $2x_1 + x_2 = 7$
$l_2 : 2x_1 + 3x_2 = 15$
$l_3: x_2 = 3, l_4 : x_1 = 0$ and $l_5 : x_2 = 0$
image
For C: Solving $l_2$ and $l_3$, we get $C(3, 3)$
For D : Solving $l_1$ and $l_3$, we get $D(2, 3)$
Shaded portion $ABCD$ is the feasible region where
$A(3.5,0), B(7.5,0), C(3, 3), D(2, 3)$
Now, minimize $Z= 4x_1 + 5x_2$
$Z$ at $A(3.5,0) = 4(3.5) + 5(0) = 14$
$Z$ at $B(7.5 0) = 4(7.5) + 5(0) = 30$
$Z$ at $C(3, 3) = 4(3) + 5(3) = 27$
$Z$ at $D(2, 3) = 4(2) + 5(3) = 23$
Thus, $Z$ is minimized at $A(3.5,0)$ and its minimum value is $14$.