The corner points of the feasible region determined by the following system of linear inequalities.
$x+3y≤60,$
$x+y ≥ 10, x, y ≥ 0$
are (0, 10), (0, 20), (60, 0) and (10, 0). Let $Z=3x+5y$, then Max Z occurs at :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (60, 0)
Objective function, $Z=3x+5y$
Maximum and Minimum occurs only at the corner points of the feasible region.
$Z_{max}=Z(60,0)$
$=3×60+5×0=180$