The point which provides the optimal solution of the linear programming problem
maximize $z =21x+35y$
$3x + 2y ≤ 30$
$4x+5y ≤ 60$
$x≥0,y≥0$
has the coordinates

$(0,12)$

The correct answer is Option (2) → $(0,12)$ **

Feasible corner points: $(0,0),\;(10,0),\;\big(\frac{30}{7},\frac{60}{7}\big),\;(0,12)$.

Evaluate $z=21x+35y$ at each point:

$z(0,0)=0$

$z(10,0)=21\cdot 10=210$

$z\big(\frac{30}{7},\frac{60}{7}\big)=21\cdot \frac{30}{7}+35\cdot \frac{60}{7}=\frac{630+2100}{7}=\frac{2730}{7}=390$

$z(0,12)=35\cdot 12=420$

Maximum value occurs at $(0,12)$ with $z=420$. Correct point: $(0,12)$.