Question

The maximum value of $z = 5x + 3y$, subjected to the conditions $3x + 5y \le 15, 5x + 2y \le 10, x, y \ge 0$ is

• A. $\frac{235}{19}$
• B. $\frac{325}{19}$
• C. $\frac{523}{19}$
• D. $\frac{532}{19}$

Answer 1

Answer: (A)

$\frac{235}{19}$

Answer 2

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 $\frac{235}{19}.$