Objective function $z=200x+500y,$ subject to the constraints $x+2y≥10, 3x+4y≤24, x≥0, y≥0,$ the minimum value of z is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2300
$x+2y≥10, 3x+4y≤24, x,y≥0$
finding intersection of $x+2y=10$ ...(1)
$3x+4y=24$ ...(2)
eq. (2) - 2 × eq. (1)
$3x+4y-2x-4y=24-20$
$x=4$
from (1) $y=3$
$z=200x+500y$
checking at corner points
$Z_A=3000$
$Z_B=2500$
$Z_C=2300$ → $Z_{min}$