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Q.
The maximum value of $z = 5x + 3y$, subjected to the conditions $3x + 5y \le 15, 5x + 2y \le 10, x, y \ge 0$ is
Linear Programming
Solution:
Given, inequalities are $3x + 5y \le 15, 5x + 2y \le 10, x, y \ge 0$
Also, given $z = 5x + 3y$
At point $A \left(2, 0\right)$
$z = 5 × 2 + 0 = 10$
At point $B\left(\frac{20}{19}, \frac{45}{19}\right)$,
$z = \frac{5\times20}{19}+\frac{3\times45}{19} = \frac{235}{19}$
At point $C \left(0, 3\right)$
$z = 5 \left(0\right) + 3 × 3 = 9$
Hence, maximum value of z is $\frac{235}{19}.$
