Q.A toy company manufactures two types of toys, A and B. Each toy A needs 2 hours of cutting and 1 hour of assembling; each toy B needs 1 hour of cutting and 2 hours of assembling. At most 120 hours of cutting time and 90 hours of assembling time are available per week. The profit is Rs 50 on each toy A and Rs 40 on each toy B. Formulate this as a linear programming problem and solve it graphically to find the maximum weekly profit.

Formulate with $x,y$ = toy A, toy B made per week, maximise $Z=50x+40y$ subject to the cutting and assembling limits, then evaluate $Z$ at every corner point of the feasible region.

Formulation. Let $x$ = number of toy A, $y$ = number of toy B made per week. Maximise

Cutting: $2x+y\le120$. Assembling: $x+2y\le90$. Non-negativity: $x\ge0,\ y\ge0$.

Corner points. On the x-axis ($y=0$): $2x\le120\Rightarrow x\le60$; $x\le90$. The smaller governs: $x=60$, giving $(60,0)$ (checking $x+2y=60\le90$, true).

On the y-axis ($x=0$): $y\le120$; $2y\le90\Rightarrow y\le45$. The smaller governs: $y=45$, giving $(0,45)$ (checking $2x+y=45\le120$, true).

Intersection of the two lines. Solving $2x+y=120$ and $x+2y=90$: from the first, $y=120-2x$; substituting,

so the lines meet at $(50,20)$, the fourth corner point. …

As in Worked Example 5, the temptation is to stop at the two axis-intercept corners and report $A(60,0)$'s Rs 3000 as the answer — the true maximum, Rs 3300, occurs strictly inside at $B(50,2 …