MHTCET 18 Apr 2026 Shift 1 Paper

Concept:In a linear programming problem, the maximum value of the objective function occurs at a corner point of the feasible region.Explanation:The feasible region is bounded by the constraint lines $3x+y=15$ and $3x+4y=24$, along with the axes $x=0$ and $y=0$.First, determine all the corner points of the feasible region.The origin is $(0,0)$.On the x-axis, put $y=0$.Then $3x \leq 15$, so $x \leq 5$.This gives the corner point $(5,0)$.On the y-axis, put $x=0$.Then $4y \leq 24$, so $y \leq 6$.This gives the corner point $(0,6)$.Next, find the intersection of the two boundary lines.Subtract $3x+y=15$ from $3x+4y=24$.This gives $3y=9$, so $y=3$.Substitute $y=3$ into $3x+y=15$.Then $3x+3=15$, so $3x=12$ and $x=4$.Thus, the intersection point is $(4,3)$.Now evaluate $Z=4x+5y$ at every corner point.At $(0,0)$: $Z=4(0)+5(0)=0$At $(5,0)$: $Z=4(5)+5(0)=20$At $(4,3)$: $Z=4(4)+5(3)=16+15=31$At $(0,6)$: $Z=4(0)+5(6)=30$The largest value among these is $31$.Answer:The maximum value of $Z$ is $31$, which matches option A.