Strategic Cost Management · Business Application of Maxima and Minima

Maxima and Minima: Basic Concepts and Conditions

Updated 11 October 2026 · Fact-checked

A maximum or minimum is the highest or lowest value of a function. To find it, differentiate, set the first derivative to zero to get stationary points, then check the second derivative. If it is negative you have a maximum; if positive, a minimum. Then put the value back into the function.

Understand Maxima and Minima: Basic Concepts

A function links one variable to another. In business, profit depends on output, and cost depends on batch size. You write this as y = f(x). The question is: at what value of x is y the highest (profit) or the lowest (cost)?

The first derivative, dy/dx, gives the slope of the curve at any point. At the top of a hill or the bottom of a valley, the curve is flat for an instant. The slope there is zero. So you set dy/dx = 0 and solve for x. The points you get are called stationary points.

A flat point can be a peak, a valley, or neither. The second derivative, d²y/dx², tells you which. It shows how the slope is changing. If the slope falls from positive to negative as you cross the point, the curve is bending down: a maximum. If the slope rises from negative to positive, the curve is bending up: a minimum.

In SCM, you use this to find the output that maximises profit, the price that maximises revenue, or the quantity that minimises cost. The method is the same each time. Only the function changes.

A maximum or minimum found this way is a local one. For an exam function such as a quadratic profit or a U-shaped cost curve, it is also the best value. If the question restricts x to a range, check the end values too.

Key rules to remember

Function: y = f(x); Define the business quantity (profit, cost, revenue) as a function of one decision variable.
First order (necessary) condition: dy/dx = f′(x) = 0; Solve for x to get stationary points. This alone does not tell you maximum or minimum.
Second order condition for a maximum: d²y/dx² < 0 at the stationary point; The curve bends downward. Profit-type functions usually show this.
Second order condition for a minimum: d²y/dx² > 0 at the stationary point; The curve bends upward. Cost-type functions usually show this.
Inconclusive case: d²y/dx² = 0; The test fails. Check the sign of dy/dx on both sides of the point.
Power rule: d(axⁿ)/dx = a·n·xⁿ⁻¹; The constant term differentiates to zero.
Revenue and profit link: Profit = Revenue − Cost; at the optimum, MR = MC; Marginal revenue is dR/dx and marginal cost is dC/dx. This follows from setting d(Profit)/dx = 0.

How to solve Maxima and Minima: Basic Concepts questions

Use this method for any single-variable maxima and minima question.

1Write the function to be optimised in terms of one variable. If profit is asked, build Profit = Revenue − Cost first.
2Differentiate once to get dy/dx.
3Set dy/dx = 0 and solve for x. List every stationary point.
4Differentiate again to get d²y/dx².
5Substitute each stationary point into d²y/dx². Negative means maximum; positive means minimum.
6Reject values that make no business sense, such as negative output or output above capacity.
7Substitute the chosen x into the original function to find the maximum profit or minimum cost.
8State the answer with units and a one-line recommendation.

Quickest way: Differentiate twice and read the sign

When to use it: Use it when the function is a polynomial, such as a quadratic or cubic, and the question asks for the optimal level and its value.

Differentiate term by term: multiply by the power, reduce the power by one.
Solve the first derivative equal to zero. For a quadratic profit this is a linear equation.
For a quadratic, the second derivative is a constant, so read its sign directly: negative for a maximum, positive for a minimum.
Put x back in the original function, not the derivative, to get the value.
Check that x is feasible before you write the answer.

Common mistakes in Maxima and Minima: Basic Concepts

Stopping after dy/dx = 0 and calling the point a maximum or minimum.
Students treat a stationary point as the answer.
Fix: Always take the second derivative and state its sign. Marks are given for the condition.
Substituting x into the derivative to find the maximum profit.
Students confuse the derivative with the original function.
Fix: The derivative gives the slope, which is zero there. Use the original function for the value.
Reversing the signs: calling d²y/dx² > 0 a maximum.
Rote memory without the picture of the curve.
Fix: Remember: negative bends down like a hill (maximum); positive bends up like a bowl (minimum).
Differentiating errors, such as forgetting that a constant becomes zero or that the fixed cost disappears.
Rushing under time pressure.
Fix: Differentiate each term separately. Fixed cost drops out in the derivative but stays in the profit value.
Accepting an infeasible stationary point, such as negative units.
The algebra gives two roots and students keep both.
Fix: Test each root with the second derivative and the business limits, then keep only the valid one.
Treating d²y/dx² = 0 as proof of neither maximum nor minimum.
Students over-apply the rule.
Fix: A zero second derivative is inconclusive. Check the sign of the first derivative on either side of the point.

Worked examples

Example 1

A firm's profit function is P = −2x² + 120x − 700, where x is units produced in hundreds and P is in ₹ thousand. Find the output that maximises profit and the maximum profit.

Show the solution

dP/dx = −4x + 120.
Set −4x + 120 = 0, so x = 30.
d²P/dx² = −4, which is less than 0. So x = 30 gives a maximum.
P at x = 30 = −2(900) + 120(30) − 700 = −1,800 + 3,600 − 700 = 1,100.
Units: x = 30 hundred, that is 3,000 units. Profit = ₹1,100 thousand = ₹11,00,000.

Answer: Profit is maximised at 3,000 units, and the maximum profit is ₹11,00,000.

Example 2

The total cost of a batch of x units is C = x³ − 12x² + 60x + 200 (₹ in thousand), for x > 0. Find the batch size at which total cost is at a stationary point, test whether it is a minimum, and state the cost there.

Show the solution

dC/dx = 3x² − 24x + 60.
Set 3x² − 24x + 60 = 0. Divide by 3: x² − 8x + 20 = 0.
Discriminant = 64 − 80 = −16, which is negative. There is no real root.
So dC/dx is always positive: it has no stationary point. Total cost keeps rising with x.
Conclusion: no interior minimum exists. The lowest cost in the range x > 0 is approached as x tends to 0, where cost tends to ₹200 thousand (fixed cost).

Answer: There is no stationary point, so there is no batch size with a maximum or minimum of total cost. Cost rises with output, and a question like this would normally ask about average cost instead.

Exam tips

Show all three parts: first derivative set to zero, second derivative sign, and the final value. Each carries marks in written answers.
In MCQs, check the sign of the second derivative first. It often eliminates two options at once.
Read units carefully. Many questions give x in hundreds or ₹ in thousands, and the final answer must be converted.
Write one line of business meaning, such as produce 3,000 units to earn the highest profit. SCM questions reward a clear recommendation.
If the problem gives capacity or a minimum order, check the feasible range before you finalise x.

Practice questions from Business Application of Maxima and Minima

Maxima and Minima: Basic Concepts in other exams

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

CMA FoundationMaxima and Minima

Maxima and Minima: Basic Concepts: frequently asked questions

What is the first order condition for a maximum or minimum?

It is dy/dx = 0. This means the slope of the function is zero at the point. It is necessary for a maximum or minimum but does not tell you which one.

How do I know if a stationary point is a maximum or minimum?

Find the second derivative and put the stationary point in it. If the value is negative, it is a maximum. If it is positive, it is a minimum.

What if the second derivative is zero?

The test is inconclusive. Check the sign of the first derivative just before and just after the point. A change from positive to negative means a maximum, and from negative to positive means a minimum.

Where is this used in CMA Final SCM?

It is used to find the profit-maximising output or price, the cost-minimising quantity, and the point where marginal revenue equals marginal cost. It is the base for the later topics in this chapter.