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 :
Answer & explanation
Correct answer: option 2
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)
$25x + 10y-6x-10y=50-30$
$19x=20⇒x=\frac{20}{19}$
$⇒x=\frac{45}{19}$
| corner points | value of Z |
| $A(0,0)$ | $Z_A=0$ |
| $B(0,3)$ | $Z_B=9$ |
| $C(2,0)$ | $Z_C=10$ |
| $D(\frac{20}{19},\frac{45}{19})$ | $Z_D=\frac{235}{19}$ |
Maximum value - minimum value = $\frac{235}{19}$