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 :

2300

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

$Z_A=3000$

$Z_B=2500$

$Z_C=2300$ → $Z_{min}$