Q. The maximum value of $z = 9x + 11y$ subject to $3x+2y \, \le\,12, 2x+3y\, \le\,12, x\, \ge\,0, y\ge\,0$ is $\ldots\ldots$
MHT CETMHT CET 2019
Solution:
We have, $Z=9 x+11 y$ Subject to $3 x+2 y \leq 12,2 x+3 y \leq 12$ On taking given constraints as equation, we get the following graph.
Intersection point of lines $3 x+2 y=12$ and $2 x+3 y=12$ is $C\left(\frac{12}{5}, \frac{12}{5}\right)$
Here, $O A C B O$ is the feasible region whose corner points are $O(0,0), A(4,0),C\left(\frac{12}{5}, \frac{12}{5}\right)$ and $B(0,4)$.
Corner points
$Z = 9x + 11y$
$O(0,0)$
$0+0=0$
$A(4,0)$
$9 \times 4+11 \times 0=36$
$B(0,4)$
$9 \times 0+11 \times 4=44$
$C\left(\frac{12}{5}, \frac{12}{5}\right)$
$9 \times \frac{12}{5}+11 \times \frac{12}{5}=48$ (maximum)
| Corner points | $Z = 9x + 11y$ |
| $O(0,0)$ | $0+0=0$ |
| $A(4,0)$ | $9 \times 4+11 \times 0=36$ |
| $B(0,4)$ | $9 \times 0+11 \times 4=44$ |
| $C\left(\frac{12}{5}, \frac{12}{5}\right)$ | $9 \times \frac{12}{5}+11 \times \frac{12}{5}=48$ (maximum) |