Q.A company manufactures two products, X and Y, using two machines. Product X requires 3 hours on machine $M_1$ and 1 hour on machine $M_2$; product Y requires 4 hours on $M_1$ and 2 hours on $M_2$. Machine $M_1$ is available for at most 24 hours and $M_2$ for at most 10 hours per week. The profit is Rs 20 per unit of X and Rs 30 per unit of Y. How many units of each product should be produced weekly to maximize profit?

Let $x=$ units of X and $y=$ units of Y produced weekly.

At $y=0$: $M_1$ gives $x\le8$, $M_2$ gives $x\le10$; binding $x\le8$, vertex $(8,0)$ — check $M_2$: $8\le10$, slack.

At $x=0$: $M_1$ gives $y\le6$, $M_2$ gives $y\le5$; binding $y\le5$, vertex $(0,5)$ — check $M_1$: $20\le24$, slack.

Intersection: $3x+4y=24,\ x+2y=10$. From the second, $x=10-2y$; substituting, $3(10-2y)+4y=24\Rightarrow30-6y+4y=24\Rightarrow-2y=-6\Rightarrow y=3,\ x=10-6=4$, giving $(4,3)$.

Corners: $(0,0),\ (8,0),\ (4,3),\ (0,5)$.

Maximum $=170$, at $(4,3)$ — i.e. $4$ units of X and $3$ units of Y, weekly profit Rs 170.

Standard corner-point method with two machine-hour constraints; find all four corners and evaluate the profit function at each.

Mixing up which machine's hour-limit belongs to which product when writing the constraints, which would give the wrong slope for the boundary lines and hence the wrong intersection point.