Consider the following L.L.P.
Minimize $z = 30x-30y+ 1800$;
subject to $x + y ≤ 30,x≤15,y ≤ 20, x + y ≥ 15$ and $x, y ≥0$.
Then it attains its optimal value at the point

(0, 20)

The correct answer is Option (1) → (0, 20)

Feasible vertices: \((0,15),(0,20),(10,20),(15,0),(15,15)\)

Objective: \(z=30x-30y+1800\)

Values: \(z(0,15)=1350,\ z(0,20)=1200,\ z(10,20)=1500,\ z(15,0)=2250,\ z(15,15)=1800\)

Minimum value \(=1200\) at the point \({(0,20)}\)