MHTCET 13 May 2023 Shift 2 Solved Paper

We have, z=10x+25y
The shaded area in the given figure is a solution system set for some system of inequations.
The constraints are x≤3,y≤3,x+y≤5,x≥0 and y≥0.
The vertices of the feasible region are (0,0),(0,3),(2,3),(3,2) and (3,0)
At (0,0),‌‌z=10×0+25×0=0
At A(0,3),‌‌z=10(0)+25(3)=75
At B(2,3),‌‌z=10(2)+25(3)=95
At C(3,2),‌‌z=10(3)+25(2)=80
At D(3,0),‌‌z=10(3)+25(0)=30
∴ maximum value of z=95, where x=2 and y=3.