MHTCET 17 Apr 2026 Shift 1 Paper

Concept:Since $z=5x+10y=5(x+2y)$, maximizing $z$ is equivalent to maximizing $x+2y$, which is directly limited by the constraint $x+2y \le 10$.Explanation:The constraint $x+2y \le 10$ gives the maximum possible value of $x+2y$ as $10$.Hence, the maximum value of $z$ is $z=5(10)=50$.This maximum is achieved only when $x+2y=10$.So we find the feasible portion of the line $x+2y=10$.On the $y$-axis, $x=0$, so $2y=10$, giving $y=5$. This gives the point $(0,5)$.Now intersect with $3x+y=12$.From $x+2y=10$, we get $x=10-2y$.Substitute into $3x+y=12$:$3(10-2y)+y=12 \Rightarrow 30-5y=12 \Rightarrow y=\frac{18}{5}$.Then $x=10-2\left(\frac{18}{5}\right)=\frac{14}{5}$.The other point is $\left(\frac{14}{5},\frac{18}{5}\right)$.Every point on the segment joining these two points satisfies $x+2y=10$ and lies in the feasible region, so the maximum value of $z$ is attained along the whole segment.Answer:The maximum value occurs at every point on the line segment joining $(0,5)$ and $\left(\frac{14}{5},\frac{18}{5}\right)$, i.e., Option D.