Difference between the maximum and minimum value of the objective function $Z=5x+3y , $ subject to the constraints : $3x+5y≤15;5x+2y ≤10: x, y ≥0$ is :

$\frac{235}{19}$

The correct answer is Option (2) → $\frac{235}{19}$

finding intersection of 

$5x+2y=10$  ...(1)

$3x+5y=15$  ...(2)

5 × eq. (1) - 2 × eq. (2)

$25x + 10y-6x-10y=50-30$

$19x=20⇒x=\frac{20}{19}$

$⇒x=\frac{45}{19}$

Maximum value - minimum value = $\frac{235}{19}$