MHTCET 11 Apr 2026 Shift 1 Paper

Concept:For a linear programming problem, the maximum value of a linear objective function always occurs at a corner point of the feasible region.Explanation:The objective function to be maximized is:$$Z = x + y$$The given constraints are:$$x + 2y \leq 70$$$$2x + y \leq 95$$$$x \geq 0, \quad y \geq 0$$First, identify the corner points of the feasible region.These are $(0,0)$, $(47.5,0)$, $(0,35)$, and the intersection point of the two boundary lines.Solve the boundary equations simultaneously:$$x + 2y = 70$$$$2x + y = 95$$Multiply the second equation by $2$:$$4x + 2y = 190$$Subtract the first equation from this result:$$(4x + 2y) - (x + 2y) = 190 - 70$$$$3x = 120$$$$x = 40$$Substitute $x = 40$ into $x + 2y = 70$:$$40 + 2y = 70$$$$2y = 30$$$$y = 15$$Thus, the intersection point is $(40, 15)$.Now evaluate $Z = x + y$ at each corner point:$(0,0)$: $Z = 0$$(47.5,0)$: $Z = 47.5$$(0,35)$: $Z = 35$$(40,15)$: $Z = 55$The highest value is $55$, which occurs at $(40,15)$.Answer:Option D: $(40,15)$