Strategic Cost Management · Business Application of Maxima and Minima

Cost, Revenue and Profit Functions in Maxima and Minima

Q: How do I derive the profit function from a demand function?

Rearrange the demand function to get price in terms of x. Multiply by x to get TR. Subtract the total cost function from TR. The result, π = TR − TC, is a function of x alone, ready to differentiate.

Q: What is the difference between average cost and marginal cost with calculus?

Average cost is TC ÷ x, the cost per unit over all units. Marginal cost is dTC/dx, the extra cost of the next unit. When MC is below AC, AC is falling. When MC is above AC, AC is rising. At minimum AC they are equal.

Q: Why does fixed cost not affect the profit-maximising output?

Fixed cost is a constant, and its derivative is zero. So it does not change the MR = MC condition. It does reduce profit, so include it when you calculate the maximum profit figure.

Q: Is MR = MC enough to prove maximum profit?

No. It is the first-order condition and may also give a minimum. You must check that d²π/dx² is negative, which means MC is rising faster than MR at that output.

Updated 11 October 2026 · Fact-checked

Cost, revenue and profit functions write total cost, total revenue and profit in terms of output x. Use the demand equation to express price in x, form TR = p·x, subtract TC to get profit, then set dπ/dx = 0 (MR = MC) and check d²π/dx² < 0 to confirm a maximum.

Understand Cost, Revenue and Profit Functions

A function here is just a rule that links output (x units) to a money amount. Costs, revenue and profit all change with x, so each can be written as an equation in x. Once you have the equation, calculus finds the best output.

Total cost (TC) = fixed cost + variable cost. Fixed cost does not change with x, so it vanishes when you differentiate. Average cost (AC) is TC ÷ x, the cost per unit. Marginal cost (MC) is dTC/dx, the extra cost of one more unit (as an approximation for small changes).

Total revenue (TR) = price × quantity. A demand function such as p = 140 − 0.03x tells you the price at which x units will sell. Put that price into TR = p·x to get TR purely in x. Average revenue (AR) is TR ÷ x, which equals price. Marginal revenue (MR) is dTR/dx. When price falls as quantity rises, MR is lower than price.

Profit (π) = TR − TC. Differentiate and set dπ/dx = 0. This is the same as MR = MC. Then check the second derivative: if d²π/dx² < 0, profit is at a maximum. The same logic applies to cost: AC is lowest where d(AC)/dx = 0, and there MC = AC.

The exam tests set-up more than calculus. Most marks go to forming the right function from the data, differentiating correctly, checking the second-order condition and stating the answer in business words.

Key rules to remember

Total cost: TC = FC + VC(x); FC is constant. It affects profit but not the profit-maximising output.
Average and marginal cost: AC = TC ÷ x ; MC = d(TC)/dx; Find AC by dividing the whole TC by x before differentiating AC.
Total, average and marginal revenue: TR = p·x ; AR = TR ÷ x = p ; MR = d(TR)/dx; Write p in terms of x from the demand function first.
Linear demand case: p = a − bx ⇒ TR = ax − bx² ; MR = a − 2bx; MR has the same intercept as demand but twice the slope. This holds only for linear demand.
Profit function: π = TR − TC; Use total figures, not per-unit figures, unless the question asks for per-unit profit.
First-order condition for maximum profit: dπ/dx = 0 ⇔ MR = MC; This gives the candidate output. It is not yet proved to be a maximum.
Second-order condition: d²π/dx² < 0 ⇒ maximum ; d²π/dx² > 0 ⇒ minimum; Equivalent to the MC curve cutting MR from below.
Minimum average cost: d(AC)/dx = 0 and d²(AC)/dx² > 0 ⇒ MC = AC; At minimum AC the marginal cost equals average cost.
Maximum revenue: MR = 0 and d²(TR)/dx² < 0; Revenue maximisation gives a different output from profit maximisation unless MC = 0.
MR and price elasticity: MR = p × (1 − 1/|e|); Here e is the price elasticity of demand. MR is positive only when |e| > 1.

How to solve Cost, Revenue and Profit Functions questions

Use this order for any cost-revenue-profit question. It keeps the working clean and shows the examiner each mark-earning step.

1Define x clearly (units produced and sold) and note any assumption, such as output equals sales.
2Write the cost function TC in terms of x. Add fixed and variable parts if they are given separately.
3Get price in terms of x from the demand function. Form TR = p·x. If the question gives a fixed price, TR = price × x.
4Form the profit function π = TR − TC. Simplify fully before differentiating.
5Differentiate: dπ/dx = 0. Show that this is the same as MR = MC. Solve for x. Reject negative or impractical roots.
6Find d²π/dx². State its sign at the solution to prove a maximum (or d²/dx² > 0 for minimum cost).
7Substitute x into the demand function for price, then into π for maximum profit. Check that MR = MC.
8State the answer in words: output, price, profit and, if asked, the recommendation.

Quickest way: Direct MR = MC shortcut

When to use it: Use it when the question asks only for the profit-maximising output, price or profit and you already have demand and cost in simple form.

Write TR = (demand price) × x and differentiate to get MR. For p = a − bx, MR = a − 2bx directly.
Differentiate TC to get MC. Fixed cost drops out.
Equate MR = MC and solve for x.
Write the second derivative of π as (slope of MR) − (slope of MC). If it is negative, it is a maximum.
Find price from the demand function. Compute profit from TR − TC with the fixed cost included. Do a quick check that MR = MC at your x.

Common mistakes in Cost, Revenue and Profit Functions

Using price p as marginal revenue.
AR and MR look alike, and they are equal only when price does not change with quantity.
Fix: If price falls as x rises, always build TR = p·x first and differentiate it. For p = a − bx, MR = a − 2bx.
Differentiating AC without dividing TC by x, or confusing AC with MC.
Both are cost per unit in students' minds.
Fix: AC = TC ÷ x. MC = dTC/dx. For minimum AC, differentiate AC and set it to zero. Then verify MC = AC.
Skipping the second-order test.
Students assume that dπ/dx = 0 always gives a maximum.
Fix: Always compute d²π/dx² and state its sign. Examiners give a separate mark for it.
Leaving fixed cost out of the final profit.
Fixed cost disappears on differentiation, so it is forgotten later.
Fix: Use the optimal x in the full π = TR − TC, with fixed cost included. Then check the final number.
Putting the optimal x into the cost or revenue function for the price.
Students mix up price, AR and TR.
Fix: Get the price from the demand function p = f(x). Then TR = p × x. Check that TR ÷ x gives the same price.
Accepting a negative or fractional output without comment.
Quadratic roots are copied without reading the business context.
Fix: Reject negative roots. If units must be whole, compare profit at the two nearest whole numbers.

Worked examples

Example 1

A company in Pune sells a product with demand function p = 140 − 0.03x, where p is price in ₹ per unit and x is units. Total cost is TC = 20,000 + 40x + 0.01x². Find the output that maximises profit, the price, and the maximum profit.

Show the solution

TR = p·x = 140x − 0.03x².
π = TR − TC = 140x − 0.03x² − 20,000 − 40x − 0.01x² = 100x − 0.04x² − 20,000.
dπ/dx = 100 − 0.08x = 0, so x = 100 ÷ 0.08 = 1,250 units.
d²π/dx² = −0.08 < 0, so profit is a maximum.
Price = 140 − 0.03 × 1,250 = 140 − 37.5 = ₹102.50 per unit.
Profit = 100 × 1,250 − 0.04 × (1,250)² − 20,000 = 1,25,000 − 62,500 − 20,000 = ₹42,500.
Check: MR = 140 − 0.06 × 1,250 = 65. MC = 40 + 0.02 × 1,250 = 65. So MR = MC.

Answer: Produce and sell 1,250 units at ₹102.50 per unit. Maximum profit is ₹42,500.

Example 2

A firm's total cost is TC = 0.1x² + 10x + 4,000 (₹). (a) Find the output at which average cost is minimum and the minimum AC. (b) Show that MC = AC there. (c) If the selling price is fixed at ₹60 per unit, find the profit at that output, and the output at which profit would be highest.

Show the solution

AC = TC ÷ x = 0.1x + 10 + 4,000/x.
d(AC)/dx = 0.1 − 4,000/x² = 0, so x² = 40,000 and x = 200 units.
d²(AC)/dx² = 8,000/x³ > 0 for x > 0, so AC is a minimum.
Minimum AC = 0.1 × 200 + 10 + 4,000 ÷ 200 = 20 + 10 + 20 = ₹50 per unit.
MC = d(TC)/dx = 0.2x + 10. At x = 200, MC = 40 + 10 = ₹50, which equals AC.
At x = 200: TR = 60 × 200 = ₹12,000. TC = 0.1 × 40,000 + 2,000 + 4,000 = ₹10,000. Profit = ₹2,000.
For maximum profit with fixed price, MR = 60. Set MC = MR: 0.2x + 10 = 60, so x = 250.
At x = 250: TR = ₹15,000. TC = 0.1 × 62,500 + 2,500 + 4,000 = 6,250 + 2,500 + 4,000 = ₹12,750. Profit = ₹2,250.
d²π/dx² = −0.2 < 0, so x = 250 is a maximum.

Answer: Minimum AC is ₹50 at 200 units, where MC = AC = ₹50. Profit at 200 units is ₹2,000. Profit is highest at 250 units, at ₹2,250, so minimum average cost output is not the profit-maximising output.

Exam tips

In MCQs, check the formula first: MR = d(p·x)/dx, not p. Many wrong options are built from using price as MR.
In written answers, show the second-order test as its own line with the sign. It is an easy mark to protect.
Put the optimal x back into MR and MC as a check. If they differ, find the error before you write the answer.
Read what is asked: maximum revenue (MR = 0), minimum average cost (MC = AC) and maximum profit (MR = MC) are three different problems.
End with a one-line recommendation in business terms, such as the output to produce and the price to charge. Case-based questions reward this.

Practice questions from Business Application of Maxima and Minima

Cost, Revenue and Profit Functions in other exams

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

Cost, Revenue and Profit Functions: frequently asked questions

How do I derive the profit function from a demand function?

Rearrange the demand function to get price in terms of x. Multiply by x to get TR. Subtract the total cost function from TR. The result, π = TR − TC, is a function of x alone, ready to differentiate.

What is the difference between average cost and marginal cost with calculus?

Average cost is TC ÷ x, the cost per unit over all units. Marginal cost is dTC/dx, the extra cost of the next unit. When MC is below AC, AC is falling. When MC is above AC, AC is rising. At minimum AC they are equal.

Why does fixed cost not affect the profit-maximising output?

Fixed cost is a constant, and its derivative is zero. So it does not change the MR = MC condition. It does reduce profit, so include it when you calculate the maximum profit figure.

Is MR = MC enough to prove maximum profit?

No. It is the first-order condition and may also give a minimum. You must check that d²π/dx² is negative, which means MC is rising faster than MR at that output.