Q44

2 marksSection E

(a) How many bags of each type should the man buy to get maximum profit ?

OR

(b) Find the profit that the man can earn by selling all the bags.

Linear Programming

LPP — Optimal Solution by Corner-Point Method

Official Answer

Optimise $Z = 100x + 90y$ over the feasible region formed by the cost, storage and non-negativity constraints.

The LPP:

Maximise $Z = 100x + 90y$
subject to $3x + 2y \le 25$ (cost), $x + y \le 10$ (storage), $x \ge 0$ , $y \ge 0$ .

Corner points and Z-values:

$(0, 0)$ : $Z = 0$
$(25/3, 0) \approx (8.33, 0)$ : $Z \approx 833$
$(5, 5)$ : $Z = 500 + 450 = 950$
$(0, 10)$ : $Z = 900$

(a) Bags to buy for maximum profit:

Maximum Z occurs at $(5, 5)$ , so buy 5 bags of rice and 5 bags of wheat.

(b) Maximum profit on selling all bags:

Maximum profit = Rs 950.

corner point methodfeasible regionmaximise Z = 100x + 90yvertex (5, 5)5 bags rice 5 bags wheatmaximum profit Rs 9503x + 2y ≤ 25x + y ≤ 10

Marking Scheme

11 mark: correctly identifying the corner points of the feasible region, including the vertex $(5, 5)$ from solving $3x + 2y = 25$ and $x + y = 10$ .
21 mark: evaluating Z at the corners and concluding the maximum Z = Rs 950 at $(5, 5)$ ; for (a) state $x = 5, y = 5$ , for (b) state profit = Rs 950.
3Full 2 marks for whichever alternative (a) or (b) is attempted with correct optimum.

Hint

List the corner points of the region bounded by $3x + 2y \le 25$ , $x + y \le 10$ , $x, y \ge 0$ ; the vertex $(5, 5)$ comes from solving the two lines simultaneously. Evaluate $Z = 100x + 90y$ at each.

Quick Oral Answer

Evaluating $Z = 100x + 90y$ at the corners, the maximum is Rs 950 at $(5, 5)$ , so the man should buy 5 bags of rice and 5 bags of wheat to earn the maximum profit of Rs 950.

Analysis & Explanation

This part solves the LPP formulated in Q38(i)-(ii) by the corner-point (extreme-point) method.

Concept:

The feasible region is a bounded convex polygon; the optimal value of a linear objective always occurs at one of its corner (vertex) points, so we only need to evaluate Z there.
The key vertex $(5, 5)$ is the intersection of the two active lines $3x + 2y = 25$ and $x + y = 10$ .

Working the intersection:

From $x + y = 10$ , $y = 10 - x$ ; substituting, $3x + 2(10 - x) = 25 \to x + 20 = 25 \to x = 5, y = 5$ .

Exam trap:

Do not stop at the axis point $(8.33, 0)$ ; although it satisfies the budget, it gives only Rs 833, less than Rs 950. Every vertex must be tested.
Both alternatives (a) and (b) lead to the same optimum — (a) asks for the bag mix, (b) asks for the resulting profit; state $x = 5, y = 5$ and Z = Rs 950 to be safe.

Real-world:

This is a classic budget-and-capacity trade-off a trader faces: an equal 5-5 split beats loading up on the higher-cost rice alone.

Common Mistakes

1Testing only the axis intercept $(8.33, 0)$ and reporting Rs 833 instead of checking the interior vertex $(5, 5)$ that gives Rs 950.
2Arithmetic slip while solving $3x + 2y = 25$ with $x + y = 10$ — forgetting to substitute $y = 10 - x$ correctly gives a wrong vertex.
3Overlooking $(0, 10)$ as a feasible corner because they assume the budget line is binding there, when in fact $2(10) = 20 \le 25$ .

Interesting Facts

The corner-point method rests on the Fundamental Theorem of Linear Programming: an optimum, if it exists over a bounded feasible region, is always attained at a vertex.

Here two constraints are simultaneously binding at the optimum (budget fully or partly used, storage fully used at $5+5 = 10$ bags), which is typical of an efficient LPP solution.

At $(5, 5)$ the man spends $1800(5) + 1200(5)$ = Rs 15,000 exactly and uses all 10 storage slots — both resources are fully utilised.

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Frequently Asked Questions

Why is the maximum profit found at $(5, 5)$ and not at the axis point $(8.33, 0)$ ?

The corner-point method requires testing every vertex. At $(8.33, 0)$ , $Z = 833$ , but at $(5, 5)$ , $Z = 500 + 450 = 950$ , which is larger. Since the optimum of a linear objective on a bounded region lies at a vertex, $(5, 5)$ gives the true maximum profit of Rs 950.

How do I get the vertex $(5, 5)$ ?

It is the intersection of the two binding lines $3x + 2y = 25$ and $x + y = 10$ . Substituting $y = 10 - x$ into the first: $3x + 2(10 - x) = 25 \to x + 20 = 25 \to x = 5$ , so $y = 5$ .

Do parts (a) and (b) have different answers?

No — they describe the same optimum. Part (a) asks for the bag mix (5 rice, 5 wheat) and part (b) asks for the resulting profit (Rs 950). Solving the LPP once answers both.