A textile shop manufactures shirts and trousers. The profit per shirt is ₹120 and per trouser is ₹200. The shop has at most 100 meters of fabric and 80 labour hours available per day. Each shirt requires 2 meters of fabric and 1 labour hour, while each trouser requires 4 meters of fabric and 2 labour hours. Using the graphical method, determine the number of shirts and trousers that should be produced each day to maximise profit, and state the maximum profit.

Step-by-step solution

1. Write the decision variables: let $x$ = number of shirts, $y$ = number of trousers.
2. Form the profit function: $Z=120x+200y$ to be maximised.
3. Write the resource constraints as inequalities: $2x+4y\le 100$ (fabric), $x+2y\le 80$ (labour), $x\ge 0$, $y\ge 0$.
4. Divide the fabric constraint by 2 to simplify: $x+2y\le 50$.
5. Plot the two lines $x+2y=50$ and $x+2y=80$ on the $xy$‑plane; the feasible region is bounded by $x+2y\le 50$, $x+2y\le 80$, and the axes.
6. Identify the corner points of the feasible region: (0,0), (0,25), (20,15), and (50,0).
7. Evaluate the profit $Z$ at each corner point:
8. At (0,0): $Z=0$.
9. At (0,25): $Z=120(0)+200(25)=5000$.
10. At (20,15): $Z=120(20)+200(15)=2400+3000=5400$.
11. At (50,0): $Z=120(50)+200(0)=6000$ but this point violates the labour constraint $x+2y\le 80$ (since $50+0>80$), so it is not feasible.
12. Select the feasible corner with the highest profit: (20,15) gives $Z=5400$.
13. State the optimal production plan and profit.

Concept tested: Graphical solution of two‑variable LP

Published 9 September 2026 · How this solution was prepared and checked