Strategic Cost Management · Business Application of Maxima and Minima

Profit Maximisation and Cost Minimisation Using Calculus

Updated 11 October 2026 · Fact-checked

Profit maximisation using calculus means finding the output where profit is highest. Write profit as revenue minus cost, differentiate with respect to output, and set the result to zero. This gives MR = MC. Check that the second derivative is negative, then find price from the demand function. For minimum average cost, set d(AC)/dQ = 0.

Understand Profit Maximisation and Cost Minimisation

A firm chooses an output level Q. Revenue, cost and profit all change with Q. Calculus lets you find the exact Q where profit is highest or where cost per unit is lowest, instead of testing many values.

Profit maximisation. Profit is total revenue minus total cost: π = TR − TC. Marginal revenue (MR) is the extra revenue from one more unit. Marginal cost (MC) is the extra cost. While MR is more than MC, each extra unit adds to profit. Once MC is more than MR, each extra unit reduces profit. Profit peaks where MR = MC. This is the same as setting dπ/dQ = 0.

The second-order check. A zero slope can be a peak or a valley. For a maximum, the second derivative of profit must be negative: d²π/dQ² < 0. This means MR is rising more slowly than MC, or falling faster. Always show this check. Examiners give marks for it.

Price and demand. If the firm faces a demand function such as P = a − bQ, then TR = PQ = aQ − bQ². Differentiate TR to get MR. Solve MR = MC for Q, then put Q back into the demand function to get the profit-maximising price. Do not read price from MC.

Cost minimisation. Average cost is AC = TC ÷ Q. It first falls and then rises, so it has a lowest point. Set d(AC)/dQ = 0 and solve for Q. Check that d²(AC)/dQ² > 0 for a minimum. At that output, MC equals AC. You can use this as a quick check on your answer. Note that the output with minimum average cost is usually not the output with maximum profit.

Key rules to remember

Profit function: π = TR − TC, where TR = P × Q; Write TC and TR in terms of Q before differentiating.
Marginal revenue and marginal cost: MR = d(TR)/dQ; MC = d(TC)/dQ; MC is the derivative of total cost, so fixed cost drops out of MC.
First-order condition for maximum profit: dπ/dQ = 0, which gives MR = MC; This finds the candidate output.
Second-order condition for maximum profit: d²π/dQ² < 0 (equivalently, slope of MR < slope of MC); Without this check you have not proved a maximum.
Linear demand and its MR: P = a − bQ gives TR = aQ − bQ² and MR = a − 2bQ; MR falls twice as fast as price for a linear demand curve.
Average cost and its minimum: AC = TC ÷ Q; minimum where d(AC)/dQ = 0 and d²(AC)/dQ² > 0; At this point MC = AC.
Per-unit tax or royalty: New TC = old TC + t × Q, so new MC = old MC + t; A lump-sum fixed charge adds to TC but does not change MC, so the profit-maximising output stays the same.

How to solve Profit Maximisation and Cost Minimisation questions

Use this order for any question on profit maximisation or cost minimisation. It works for both linear and non-linear functions.

1Read the question and note what is asked: output, price, maximum profit, minimum AC, or a combination.
2Write the functions given: demand (or price), TC, or AC. Convert demand into TR = P × Q if needed.
3Form the function to optimise: π = TR − TC for profit, or AC = TC ÷ Q for average cost.
4Differentiate once and set the derivative to zero. For profit, this is the same as MR = MC. Solve for Q. If the equation is quadratic, reject negative or impractical roots.
5Differentiate again and check the sign: negative for maximum profit, positive for minimum cost. Write the check in your answer.
6Find the other values asked for: price from the demand function, TR, TC, profit, or minimum AC. Substitute Q into the original functions, not the derivatives.
7Verify quickly, for example MC = AC at minimum AC, or profit = TR − TC.
8State the conclusion in a sentence with units and rupees, such as 'The firm should produce 75 units at ₹350 per unit.'

Quickest way: MR = MC shortcut with a one-line check

When to use it: Use it when demand is linear and TC is a polynomial of degree two or three, and the question asks for output, price and profit.

Write MR and MC straight from TR and TC by differentiating term by term. Drop fixed cost in MC.
Equate MR = MC and solve for Q.
Find the slopes of MR and MC. If MR slope − MC slope is negative, it is a maximum. This is your second-order check.
Put Q in P = a − bQ for price. Compute profit as (P − AC) × Q or TR − TC.
For minimum AC, if TC = F + vQ + cQ², use Q = √(F ÷ c). Verify by checking MC = AC. Use this only for this cost form.

Common mistakes in Profit Maximisation and Cost Minimisation

Setting price equal to MC instead of MR equal to MC when the firm faces a downward-sloping demand curve.
Students remember P = MC from perfect competition and apply it everywhere.
Fix: If demand depends on Q, build TR = P × Q first and differentiate it. Use P = MC only if price is fixed and does not change with Q.
Skipping the second-order condition.
Students feel the answer is complete once Q is found.
Fix: Always differentiate again, state the sign, and write 'maximum' or 'minimum'. It takes one line and secures marks.
Reading price from MR or MC instead of the demand function.
After solving MR = MC, students stop at Q and substitute into the wrong equation.
Fix: Put the optimal Q into P = a − bQ. Price is always found on the demand curve.
Differentiating AC wrongly, or setting MC = 0 to find minimum average cost.
Students confuse minimising total cost with minimising cost per unit.
Fix: Divide TC by Q first, then differentiate AC. Or use the equivalent condition MC = AC. Confirm with the second derivative.
Keeping fixed cost in MC, or forgetting that a lump-sum charge does not change the optimal output.
Students differentiate carelessly or think any new cost shifts the answer.
Fix: A constant disappears on differentiation. Only per-unit costs, such as a per-unit tax, change MC. Add a lump-sum charge only when computing final profit.
Giving a negative or fractional output without comment, or ignoring capacity limits.
Quadratic roots are accepted without checking if they make business sense.
Fix: Reject negative roots. If the question gives a capacity limit or integer units, check that Q fits. State your reasoning.

Worked examples

Example 1

A monopolist faces the demand function P = 500 − 2Q and the total cost function TC = 5,000 + 50Q + Q² (in ₹). Find the output and price that maximise profit, and the maximum profit.

Show the solution

TR = P × Q = 500Q − 2Q². So MR = d(TR)/dQ = 500 − 4Q.
MC = d(TC)/dQ = 50 + 2Q.
Set MR = MC: 500 − 4Q = 50 + 2Q. So 450 = 6Q and Q = 75 units.
Second-order check: π = TR − TC = 450Q − 3Q² − 5,000. dπ/dQ = 450 − 6Q, and d²π/dQ² = −6, which is less than 0. So profit is maximum.
Price: P = 500 − 2(75) = 500 − 150 = ₹350.
TR = 75 × 350 = ₹26,250.
TC = 5,000 + 50(75) + 75² = 5,000 + 3,750 + 5,625 = ₹14,375.
Profit = 26,250 − 14,375 = ₹11,875. Cross-check using π = 450Q − 3Q² − 5,000: 33,750 − 16,875 − 5,000 = ₹11,875.

Answer: Produce 75 units and sell at ₹350 per unit. The maximum profit is ₹11,875.

Example 2

The total cost of a plant is TC = 3,200 + 10Q + 0.5Q² (in ₹), where Q is the number of units. Find the output at which average cost is minimum, the minimum average cost, and show that MC equals AC at that output.

Show the solution

AC = TC ÷ Q = 3,200/Q + 10 + 0.5Q.
d(AC)/dQ = −3,200/Q² + 0.5. Set it to zero: Q² = 3,200 ÷ 0.5 = 6,400. So Q = 80 (reject −80).
Second-order check: d²(AC)/dQ² = 6,400/Q³. At Q = 80 this is positive, so AC is minimum.
Minimum AC = 3,200/80 + 10 + 0.5(80) = 40 + 10 + 40 = ₹90 per unit.
MC = d(TC)/dQ = 10 + Q. At Q = 80, MC = ₹90, which equals AC.
Cross-check: TC at 80 units = 3,200 + 800 + 3,200 = ₹7,200, and 7,200 ÷ 80 = ₹90.

Answer: Average cost is minimum at 80 units. The minimum average cost is ₹90 per unit, and MC = AC = ₹90 at that output.

Exam tips

In the MCQ section, expect short questions: the output where MR = MC, the sign of the second derivative, or the minimum AC. Do the differentiation on rough paper and check one option by substitution.
In written answers, show the layout: functions, derivative, solution, second-order check, then price and profit. Marks are given for each step, so do not jump to the answer.
Read whether the question asks for maximum profit, maximum revenue or minimum cost. Each needs a different function to differentiate.
Where the question adds a tax, royalty or capacity limit, change MC or the constraint first and then solve again. Conclude with a clear recommendation in rupees.
Recheck arithmetic by computing profit two ways, as TR − TC and as (P − AC) × Q, before writing the final answer.

Practice questions from Business Application of Maxima and Minima

Profit Maximisation and Cost Minimisation in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Profit Maximisation and Cost Minimisation: frequently asked questions

Why is profit maximum where MR = MC?

Each extra unit adds MR to revenue and MC to cost. While MR is above MC, profit rises. When MC goes above MR, profit falls, so the peak is where they are equal, provided the second-order condition holds.

Is the output with minimum average cost the same as the profit-maximising output?

Usually not. Profit-maximising output depends on demand and MR, while minimum average cost depends only on the cost function. They match only in special cases, such as when the MR = MC output happens to be where MC = AC.

How do I find the output for minimum average cost?

Divide TC by Q to get AC, differentiate AC with respect to Q, set it to zero and solve. Then check that the second derivative is positive. You can confirm the answer by checking that MC = AC at that output.

Do I need the second derivative test in every answer?

Yes, show it. The first derivative only gives a candidate point. The second derivative proves whether it is a maximum or a minimum, and examiners expect to see it in a written answer.

What happens to the optimal output if a fixed cost increases?

The optimal output does not change, because fixed cost disappears when you differentiate. Total profit falls by the amount of the extra fixed cost. A per-unit tax is different because it raises MC and so changes output.