Q.A company produces two goods A and B. Making one unit of A uses 3 units of raw material and 2 labour-hours; making one unit of B uses 1 unit of raw material and 2 labour-hours. There are 12 units of raw material and 10 labour-hours available. The profit is ₹6 per unit of A and ₹5 per unit of B. How many of each should be made to maximise profit, and what is the maximum profit?

Formulate, graph the bounded region, then read the maximum off the corner table.

Formulation. Let $x,y$ be the units of A and B.

Boundary lines (solid; origin test true):

• $3x+y=12$: intercepts $(4,0),(0,12)$.
• $x+y=5$: intercepts $(5,0),(0,5)$.

Corner points:

• $(0,0)$.
• $y=0$ in $3x+y=12$: $x=4$, $(4,0)$ (check $x+y=4\le5$ ✓).
• $x=0$ in $x+y=5$: $y=5$, $(0,5)$ (check $3x+y=5\le12$ ✓).
• Intersection of $3x+y=12$ and $x+y=5$: subtract the second from the first: $2x=7$?? Let us do it carefully: $(3x+y)-(x+y)=12-5\Rightarrow 2x=7\Rightarrow x=3.5$, $y=5-3.5=1.5$. Check both are feasible — yes. So $(3.5,1.5)$ is also a vertex.

Evaluate $Z=6x+5y$:

The largest value is $28.5$ at $(3.5,1.5)$. However, units of goods must be whole numbers; the best integer plan on the feasible region is checked among nearby feasible integer points. $(2,3)$: $3(2)+3=9\le12$ ✓, $2+3=5\le5$ ✓, $Z=12+15=27$. $(3,2)$: $3(3)+2=11\le12$ ✓, $3+2=5\le5$ ✓, $Z=18+10=28$. $(4,0)$: $Z=24$. So the best integer plan is $(3,2)$ with $Z=28$. …

Rounding the continuous optimum $(3.5,1.5)$ to the nearest point $(4,2)$ or $(2,3)$ without checking feasibility and value — $(4,2)$ violates $x+y\le5$, and $(2, …