Q. The maximum value of $z = 5x + 3y$, subjected to the conditions $3x + 5y \leq 15,\ 5x + 2y \leq 10;x,\ y \geq 0$ is
A) $\frac{235}{19}$
B) $\frac{325}{19}$
C) $\frac{523}{19}$
D) $\frac{532}{19}$
Explanation
(a)
Feasible region is $ABCOA$
[image:12_image19.jpeg]
Given, $z = 5x + 3y$
At point $A(2,\ 0)\ \ \ \ \ \ z = 5 \times 2 + 0 = 10$
At point $B\left( \frac{20}{19},\ \frac{45}{19} \right),\ \ \ z = \frac{5 \times 20}{19} + \frac{3 \times 45}{19} = \frac{235}{19}$
At point $C(0,\ 3)\ \ \ \ \ \ z = 5 \times 0 + 3 \times 3 = 9$
Hence, maximum value of $z$ is $\frac{235}{19}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.