Maximize the function $Z = 11x + 7y$,subject to the constraints:
$x \leq 3, y \leq 2, x \geq 0, y \geq 0$
$x \leq 3, y \leq 2, x \geq 0, y \geq 0$
(B) We have to maximize $Z = 11x + 7y$,subject to the constraints $x \leq 3, y \leq 2, x \geq 0, y \geq 0$.
The feasible region is a rectangle bounded by the lines $x = 0, x = 3, y = 0,$ and $y = 2$. The corner points of this region are $O(0, 0), A(3, 0), B(3, 2),$ and $C(0, 2)$.
We evaluate $Z$ at these corner points:
The maximum value of $Z$ is $47$ at the point $(3, 2)$.
The feasible region is a rectangle bounded by the lines $x = 0, x = 3, y = 0,$ and $y = 2$. The corner points of this region are $O(0, 0), A(3, 0), B(3, 2),$ and $C(0, 2)$.
We evaluate $Z$ at these corner points:
| Corner Point | Value of $Z = 11x + 7y$ |
| $O(0, 0)$ | $11(0) + 7(0) = 0$ |
| $A(3, 0)$ | $11(3) + 7(0) = 33$ |
| $B(3, 2)$ | $11(3) + 7(2) = 33 + 14 = 47$ |
| $C(0, 2)$ | $11(0) + 7(2) = 14$ |
The maximum value of $Z$ is $47$ at the point $(3, 2)$.
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