Q.A furniture company manufactures tables and chairs. Each table needs 2 hours of carpentry and 1 hour of finishing; each chair needs 1 hour of carpentry and 3 hours of finishing. The factory has at most 100 hours of carpentry and 120 hours of finishing available per week. The profit is ₹40 per table and ₹30 per chair. Formulate this as a linear programming problem to maximise weekly profit.

Identify the decision variables, write profit as the objective function, and turn the two resource limits into $\le$ constraints, adding non-negativity.

Carpentry: $2x+y\le100$. Finishing: $x+3y\le120$. Non-negativity: $x\ge0, y\ge0$.

Step 1 — Decision variables. Let $x$ = number of tables made per week, $y$ = number of chairs made per week.

Step 2 — Objective function. Each table earns ₹40 profit and each chair ₹30, so total weekly profit is $Z=40x+30y$, to be maximised.

Step 3 — Constraints. Carpentry: each table needs 2 hours, each chair 1 hour, at most 100 hours available: $2x+y\le100$. Finishing: each table needs 1 hour, each chair 3 hours, at most 120 hours available: $x+3y\le120$.

Step 4 — Non-negativity. A negative number of tables or chairs is meaningless: $x\ge0, y\ge0$.

Putting it together:

Independent check. Re-reading each constraint against the word problem once more: the carpentry constraint's coefficients ($2$ for tables, $1$ for chairs) match 'each table needs 2 hours of carpentry; each chair needs 1 hour' exactly, and the finishing constraint's coefficients ($1$ for tables, $3$ for chairs) match 'each table needs 1 hour of finishing; each chair needs 3 hours' exactly — confirming no coefficient was swapped between the two resources.

Swapping which resource each product's coefficient belongs to (writing the carpentry constraint with the finishing hours, or vice versa) is a common error — always re-check each coefficient against the specific resource named in the word problem.