Fundamentals of Business Mathematics and Statistics · Calculus - Application in Business

Maxima and Minima in Business Mathematics

Q: What is the difference between maxima and minima?

A maximum is a point where the function reaches a peak value in its neighbourhood. A minimum is a point where it reaches a lowest value. In business, maximum usually refers to profit or revenue, and minimum to cost.

Q: How do I know whether a critical point is a maximum or a minimum?

Find f''(x) and substitute the critical point. If f''(x) is negative, it is a maximum. If it is positive, it is a minimum.

Q: What if the second derivative is zero?

Then the second derivative test gives no answer. At CMA Foundation level such cases are rare. You can check the sign of f'(x) on either side of the point instead.

Q: How do I find the output that minimises average cost?

Divide total cost by x to get average cost. Differentiate it, set the derivative to zero and solve for x. Confirm with the second derivative being positive.

Updated 10 October 2026 · Fact-checked

Maxima and minima are the highest and lowest values of a function. To find them, solve f'(x) = 0 to get critical points, then check f''(x). If f''(x) < 0 the point is a maximum. If f''(x) > 0 it is a minimum. Put the x value back into f(x) for the value.

Understand Maxima and Minima

A business often wants the best possible result: the largest profit or the smallest average cost. If you write that quantity as a function of output x, the best value sits at a peak or a valley of the graph.

At a peak or valley the curve is momentarily flat. Its slope is zero. Slope is the first derivative, so at such a point f'(x) = 0. The x values where this happens are called critical points or stationary points.

Not every critical point is a peak. You must decide which type it is. Look at the curvature using the second derivative. If the curve bends downward (f''(x) < 0), the point is a maximum. If it bends upward (f''(x) > 0), the point is a minimum.

The answer has two parts: the point where it occurs (the x value) and the maximum or minimum value (f(x) at that x). Read the question carefully, because MCQs ask for one or the other. For example, profit is maximised at output 50 units, but the maximum profit is a rupee amount.

Key formulas to remember

Condition for a stationary point: f'(x) = 0; Solve this equation to get the critical points.
Test for maximum: f'(x) = 0 and f''(x) < 0; The function has a local maximum at that x.
Test for minimum: f'(x) = 0 and f''(x) > 0; The function has a local minimum at that x.
Profit: P(x) = R(x) − C(x); Maximise this. At the optimum, marginal revenue equals marginal cost (R'(x) = C'(x)) with P''(x) < 0.
Average cost: AC = C(x) ÷ x; Differentiate AC itself to minimise it, not C(x).
Power rule: d/dx (xⁿ) = n·xⁿ⁻¹; The derivative of a constant is 0.

How to solve Maxima and Minima questions

Use this method for any maxima or minima question, whether it is stated in plain algebra or as a business problem.

1Write the quantity to be optimised as a function of one variable, such as profit P(x) or average cost AC(x).
2Find the first derivative f'(x).
3Put f'(x) = 0 and solve for x. Reject values that make no sense, such as negative output.
4Find the second derivative f''(x).
5Substitute each critical x into f''(x). Negative means maximum. Positive means minimum.
6Substitute the x value into the original function to get the maximum or minimum value.
7Answer exactly what was asked: the point, the value, or both. Add units such as units or ₹.

Quickest way: Derivative, solve, sign check, then options

When to use it: Use this for MCQs where the function is a polynomial and you have about a minute per question.

Differentiate once and solve for x immediately. For a quadratic like ax² + bx + c, the critical point is x = −b ÷ 2a.
For a quadratic, skip the second derivative: a < 0 gives a maximum, a > 0 gives a minimum.
If the question asks for the value, substitute x into the original function and check which option matches.
If the options show an x value and a function value, make sure you pick the one the question wants.
For average cost, divide C(x) by x first, then differentiate.

Common mistakes in Maxima and Minima

Stopping after f'(x) = 0 and calling the answer the maximum value.
The x value is mistaken for the final answer.
Fix: Always put x back into f(x) if the question asks for the maximum or minimum value.
Skipping the second derivative test and guessing max or min.
Students assume profit means maximum and cost means minimum without checking.
Fix: Check the sign of f''(x) at the critical point, or use the sign of a for a quadratic.
Differentiating total cost to minimise average cost.
Confusing C(x) with C(x) ÷ x.
Fix: Form AC = C(x) ÷ x first, then differentiate AC and set it to zero.
Sign error with f''(x), such as treating f''(x) < 0 as a minimum.
Memorising the rule without understanding the bending of the curve.
Fix: Remember: negative second derivative bends down like a hill, so a maximum.
Forming profit as cost minus revenue.
Rushing through the setup.
Fix: Write P = R − C. Revenue is R = price × quantity.
Accepting a negative or fractional x for output.
Not checking whether the root is sensible.
Fix: Reject roots that are not possible for the situation.

Worked examples

Example 1

A firm's profit in rupees from producing x units is P(x) = −x² + 60x − 200. Find the output that maximises profit and the maximum profit.

Show the solution

P'(x) = −2x + 60.
Set P'(x) = 0: −2x + 60 = 0, so x = 30.
P''(x) = −2, which is less than 0. So x = 30 gives a maximum.
P(30) = −900 + 1,800 − 200 = 700.

Answer: Profit is maximum at 30 units, and the maximum profit is ₹700.

Example 2

The total cost of producing x units is C(x) = x³ − 12x² + 60x. Find the output at which the average cost is minimum, and the minimum average cost.

Show the solution

AC = C(x) ÷ x = x² − 12x + 60.
d(AC)/dx = 2x − 12. Set it to 0, so x = 6.
d²(AC)/dx² = 2, which is greater than 0. So x = 6 gives a minimum.
AC at x = 6: 36 − 72 + 60 = 24.

Answer: Average cost is minimum at 6 units, and the minimum average cost is ₹24 per unit.

Exam tips

Read the last line of the question. Options often include both the x value and the function value, so pick the one asked for.
For quadratic profit or cost functions, use x = −b ÷ 2a to save time.
Check your answer by substituting into the original function. A quick recalculation catches arithmetic slips.
There is no negative marking, so if time runs out, eliminate options that give negative output or impossible values and then guess.
Practise the derivative rules from Differentiation Basics and Rules until they are automatic, as errors usually come from differentiation.

Practice questions from Calculus - Application in Business

Maxima and Minima in other exams

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

CMA FinalMaxima and Minima: Basic Concepts

Maxima and Minima: frequently asked questions

What is the difference between maxima and minima?

A maximum is a point where the function reaches a peak value in its neighbourhood. A minimum is a point where it reaches a lowest value. In business, maximum usually refers to profit or revenue, and minimum to cost.

How do I know whether a critical point is a maximum or a minimum?

Find f''(x) and substitute the critical point. If f''(x) is negative, it is a maximum. If it is positive, it is a minimum.

What if the second derivative is zero?

Then the second derivative test gives no answer. At CMA Foundation level such cases are rare. You can check the sign of f'(x) on either side of the point instead.

How do I find the output that minimises average cost?

Divide total cost by x to get average cost. Differentiate it, set the derivative to zero and solve for x. Confirm with the second derivative being positive.