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 :

(60,0)

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.

$Z_{max}=Z(60,0)$

$=3×60+5×0=180$