Strategic Cost Management · Business Application of Maxima and Minima

Optimisation with Taxes, Discounts and Constraints

Q: How does a tax per unit change the profit-maximising output?

It raises marginal cost by the amount of the tax, so the condition becomes MR = MC + t. The best output falls. With linear demand and constant marginal cost, output falls by t ÷ (2b).

Q: Does a lump-sum tax change the optimal output?

No. A fixed tax does not change marginal cost or marginal revenue, so the first-order condition is the same. It only reduces total profit by the amount of the tax.

Q: How do I handle a quantity discount in a maximisation problem?

Write a separate profit function for each quantity range. Solve each one, keep only answers that lie inside their own range, and also check the range boundaries. Then choose the highest profit.

Q: How do I solve optimisation with a constraint in the exam?

If the constraint is an equation, substitute it into the profit function to leave one variable, then differentiate. If it is a limit such as capacity, first solve without it. If that answer breaks the limit, the best point is usually at the limit itself.

Updated 11 October 2026 · Fact-checked

Optimisation with taxes, discounts and constraints means finding the output or price that maximises profit or revenue after adding a real-world condition. Add the tax to cost, split discounts into ranges, or substitute the constraint. Then set the first derivative to zero, check the second derivative, and compare the results.

Understand Optimisation with Taxes, Discounts and Constraints

Basic maxima and minima ask one question: at what quantity is profit highest? You write profit as a function of output, differentiate, set the derivative to zero, and check that the second derivative is negative. This topic keeps that method and changes the function you start with.

A tax per unit raises the cost of every unit by the same amount. If the tax is ₹t per unit, total cost becomes C(q) + t·q. Marginal cost rises by t, so the best output falls. The rule is MR = MC + t. A lump-sum tax (a fixed amount, not linked to output) only reduces profit. It does not change the best output, because it does not change marginal cost.

A quantity discount makes price or cost depend on the range of quantity. You get a different function in each range. Solve each range separately, reject any answer that falls outside its range, and test the range boundaries. Then compare the profit from every candidate. The highest one wins. Do not apply calculus across the whole range at once.

A constraint is a restriction such as limited capacity, a fixed budget or a fixed total output. With one equality constraint, use it to eliminate one variable. Then you have a single-variable problem that you solve in the usual way. If the constraint is an inequality such as q ≤ 80, first find the unconstrained best. If it fits, you are done. If not, the best is usually at the limit.

Always finish with a business statement: the output, the price, the profit and what the change (tax, discount or restriction) did to them. The exam rewards the recommendation as well as the derivative.

Key rules to remember

Profit function: π = R − C; R = revenue, C = total cost. Maximise π, not R or C alone, unless the question asks for revenue or cost only.
Condition for maximum: dπ/dq = 0 and d²π/dq² < 0; Equivalent to MR = MC. The second-order check is needed to show it is a maximum.
Per-unit tax on producer: C_new(q) = C(q) + t·q, so MR = MC + t; t is the tax per unit. Use this when the producer bears the tax.
Ad valorem tax on revenue: π = (1 − k)·R − C; k is the tax rate on sales value, written as a decimal.
Lump-sum tax or fixed charge: π_new = π − T; T is a fixed amount. The optimal q does not change.
Linear demand, constant MC, per-unit tax: p = a − b·q; q* = (a − c − t) ÷ (2b); c is constant marginal cost. The price is p* = a − b·q*.
Revenue maximisation (linear demand): R = p·q = (a − bq)·q; q = a ÷ (2b); Revenue peaks where MR = 0. Profit peaks at a lower output whenever MC > 0.
Tax collected: Tax revenue = t × q*; Use the output after the tax, not the output before it.
Equality constraint: Substitute y = g(x) into π(x, y), then solve dπ/dx = 0; Reduces two variables to one.

How to solve Optimisation with Taxes, Discounts and Constraints questions

Use this sequence for any tax, discount or constraint question. It keeps the working short and shows every step an examiner looks for.

1Define the variable (q, or price p) and write down the demand relation, revenue function and cost function exactly as given.
2Bring in the variation. Add t·q to cost for a per-unit tax, subtract tax from profit for a lump sum, or write a separate function for each discount range.
3Form the profit function π = R − C (or the revenue or cost function if that is what is asked).
4Differentiate, set dπ/dq = 0 and solve. For a constraint, substitute it first so only one variable remains.
5Check d²π/dq² < 0 for a maximum (> 0 for a minimum).
6Test feasibility. The answer must be non-negative, in its discount range and within any capacity limit. If it is not, evaluate the boundary points.
7Compute price, profit and any tax collected using the final output.
8Write a one-line conclusion, for example how much output fell and how much of the tax was passed on to customers.

Quickest way: Shortcut for linear demand with a per-unit tax

When to use it: Use when demand is p = a − bq and marginal cost is constant or the cost function is linear. This is the most common exam setting.

Find marginal revenue directly: MR = a − 2bq.
Find marginal cost including tax: MC + t.
Equate them and solve: q = (a − MC − t) ÷ (2b).
Put q back into the demand function to get price, then compute profit.
Confirm the second derivative is −2b, which is negative, so it is a maximum.

Common mistakes in Optimisation with Taxes, Discounts and Constraints

Adding the tax to revenue or subtracting it from price in the wrong place, which gives a wrong MC.
Students treat the tax as a change in demand rather than a change in cost.
Fix: For a tax paid by the producer, add t·q to total cost. Then MC becomes MC + t.
Using the pre-tax output to calculate tax collected or profit after tax.
The pre-tax answer is already on the page and gets reused out of habit.
Fix: Recompute q from the new profit function. Tax collected = t × new q.
Treating a lump-sum tax like a per-unit tax and changing the optimal output.
Both are called a tax, so students change the derivative for both.
Fix: A fixed amount disappears when you differentiate. Change only the profit figure, not q.
Accepting a calculus answer that lies outside a discount range or capacity limit.
Students stop once dπ/dq = 0 gives a number.
Fix: Check the range for each answer. If it is outside, evaluate profit at the range boundaries and compare.
Skipping the second-order check or the final recommendation.
Time pressure makes students stop at the first derivative.
Fix: Write the second derivative in one line, then state the conclusion with output, price and profit.
Confusing revenue-maximising output with profit-maximising output.
Both use the same demand function, so the two problems look alike.
Fix: Revenue is maximised at MR = 0. Profit is maximised at MR = MC (plus tax). Read which one the question wants.

Worked examples

Example 1

A monopolist faces the demand function p = 120 − 2q. Total cost is C = 500 + 20q (in ₹). The government imposes a tax of ₹12 per unit on the producer. Find the profit-maximising output and price before and after the tax, the maximum profit after tax and the tax collected.

Show the solution

Before tax: R = 120q − 2q². π = 120q − 2q² − 500 − 20q = 100q − 2q² − 500.
dπ/dq = 100 − 4q = 0, so q = 25. d²π/dq² = −4 < 0, so it is a maximum.
Price = 120 − 2(25) = ₹70. Profit = 100(25) − 2(625) − 500 = 2,500 − 1,250 − 500 = ₹750.
After tax: cost = 500 + 20q + 12q = 500 + 32q. π = 120q − 2q² − 500 − 32q = 88q − 2q² − 500.
dπ/dq = 88 − 4q = 0, so q = 22. d²π/dq² = −4 < 0, so it is a maximum.
Price = 120 − 2(22) = ₹76. Profit = 88(22) − 2(484) − 500 = 1,936 − 968 − 500 = ₹468.
Tax collected = 12 × 22 = ₹264.

Answer: Before tax: output 25 units, price ₹70, profit ₹750. After tax: output 22 units, price ₹76, profit ₹468, tax collected ₹264. The price rises by ₹6, so customers bear half of the ₹12 tax and the firm bears the rest through lower profit.

Example 2

A firm makes two products, x and y (in tonnes, divisible). Profit in ₹ thousand is π = 40x + 30y − x² − y². Because of a plant restriction, total output must be exactly 20 tonnes (x + y = 20). Find the output mix that maximises profit and the maximum profit.

Show the solution

The constraint binds. Unrestricted, ∂π/∂x = 40 − 2x = 0 gives x = 20 and ∂π/∂y = 30 − 2y = 0 gives y = 15. That totals 35 tonnes, which breaks the limit of 20.
Substitute y = 20 − x: π = 40x + 30(20 − x) − x² − (20 − x)².
Expand: π = 40x + 600 − 30x − x² − (400 − 40x + x²) = 200 + 50x − 2x².
dπ/dx = 50 − 4x = 0, so x = 12.5. d²π/dx² = −4 < 0, so it is a maximum.
y = 20 − 12.5 = 7.5.
Profit = 40(12.5) + 30(7.5) − (12.5)² − (7.5)² = 500 + 225 − 156.25 − 56.25 = 512.5.

Answer: Produce 12.5 tonnes of x and 7.5 tonnes of y. Maximum profit is ₹512.5 thousand (₹5,12,500).

Exam tips

Read whether the tax is per unit, a percentage of sales or a lump sum before you write anything. Each changes the profit function differently.
In discount problems, show the range check explicitly. Marks are often given for rejecting an answer that falls outside its range.
Always compute price, profit and tax collected from the new output. Examiners look for the comparison before and after the change.
Show d²π/dq² in one line. It is quick and secures the method mark.
Finish with a recommendation or interpretation sentence, since this paper tests decision-making, not only differentiation.

Practice questions from Business Application of Maxima and Minima