Quantitative Aptitude · Differential and Integral Calculus

Maxima and Minima for CA Foundation

Updated 1 October 2026

Maxima and minima are the highest and lowest values of a function in a neighbourhood. To find them, solve f′(x) = 0 to get critical points. Then check f″(x): negative means a maximum, positive means a minimum. Put the x-value back into f(x) to get the value.

Understand Maxima and Minima

A function can rise, reach a peak, then fall. That peak is a maximum. If it falls, reaches a bottom, then rises, that bottom is a minimum. In business, you want maximum profit or revenue, and minimum cost.

The slope of the curve is f′(x). At a peak or a valley the tangent is flat, so the slope is zero. This is why you solve f′(x) = 0. The x-values you get are called critical points (or stationary points).

A zero slope alone does not tell you if it is a peak or a valley. The second derivative f″(x) tells you how the slope is changing. Near a peak, the slope goes from positive to negative, so it is decreasing and f″ is negative. Near a valley, the slope goes from negative to positive, so f″ is positive.

These are local (relative) maxima and minima. A local maximum is the highest value only near that point. It need not be the largest value of the function overall. In CA Foundation questions, the function is usually a polynomial, so you mostly work with local values. The question normally asks for either the x-value or the value of the function there, so read carefully.

Key formulas to remember

Condition for a critical point: f′(x) = 0; Solve this first. Every maximum or minimum of a smooth function is found among these x-values.
Second derivative test for maximum: f′(a) = 0 and f″(a) < 0 ⇒ maximum at x = a; The maximum value is f(a), not a.
Second derivative test for minimum: f′(a) = 0 and f″(a) > 0 ⇒ minimum at x = a; The minimum value is f(a).
First derivative test: f′ changes + to − at a ⇒ maximum; f′ changes − to + at a ⇒ minimum; Use it when f″(a) = 0, because the second derivative test then gives no result.
Profit function: Profit = Revenue − Cost, i.e. P(x) = R(x) − C(x); Maximise P(x). Equivalent condition: R′(x) = C′(x), i.e. marginal revenue = marginal cost.
Revenue function: R(x) = price × quantity = p × x; If price depends on x, substitute it first, then differentiate.
Average cost: AC = C(x) ÷ x; To minimise average cost, differentiate AC, not C.

How to solve Maxima and Minima questions

This method works for any maxima and minima question, from a plain function to a cost or profit problem.

1Write the function to be optimised as a function of one variable. For business problems, build P(x) = R(x) − C(x) or the average cost first.
2Differentiate to get f′(x).
3Put f′(x) = 0 and solve for x. These are the critical points. Reject values that make no sense, such as negative output.
4Differentiate again to get f″(x).
5Put each critical point into f″(x). If it is negative, it is a maximum. If it is positive, it is a minimum.
6If f″ equals zero at a point, use the first derivative test by checking the sign of f′ on either side.
7Put the x-value back into the original f(x) if the question asks for the maximum or minimum value.
8Re-read the question and give what it asks: x, the value, or the profit.

Quickest way: Option-based shortcut for MCQs

When to use it: Use it when the question is a polynomial and the options are numbers. It saves time and avoids a full second-derivative check.

Differentiate once and solve f′(x) = 0. Do not skip this step.
For a quadratic like ax² + bx + c, the sign of a decides the type. If a > 0 it has only a minimum, and if a < 0 only a maximum. The turning point is x = −b ÷ 2a.
For a cubic with two critical points, evaluate f″ at each point to tell which is the maximum and which is the minimum.
Eliminate options that do not match the x-value you found. If the question asks for the value, substitute into f(x) and match.
Check the question wording. Many wrong answers are the x-value when the value was asked, or the reverse.
Skip a question only if the function is long and the options give no clue. A wrong answer costs 0.25 marks.

Common mistakes in Maxima and Minima

Stopping at f′(x) = 0 and calling the answer the maximum or minimum value.
Students think solving for x is the end of the problem.
Fix: Solve f′(x) = 0 to find x. Then substitute x into f(x) if the question asks for the value.
Reversing the second derivative rule.
Positive feels like a maximum, so students mix the signs.
Fix: Remember: f″ < 0 is a frown, so a maximum. f″ > 0 is a smile, so a minimum.
Putting the critical point into f′(x) instead of f″(x) for the test.
Students rush and reuse the first derivative.
Fix: f′ at the point is already zero. The test uses f″ only.
Differentiating total cost when asked to minimise average cost.
Students do not read the function being optimised.
Fix: Write AC = C(x) ÷ x first. Differentiate AC and set it to zero.
Treating the local maximum as the largest value overall.
Students forget that a cubic can be larger or smaller elsewhere.
Fix: A local maximum is highest only near its point. A cubic is unbounded, so it takes larger values elsewhere, and the local maximum is not the overall maximum.
Giving a negative or impossible quantity as the answer.
Both roots of f′(x) = 0 are accepted without checking context.
Fix: Reject negative output and any value outside the stated range.

Worked examples

Example 1

The maximum value of f(x) = 2x³ − 15x² + 36x + 10 is: (a) 38 (b) 37 (c) 10 (d) 28

Show the solution

f′(x) = 6x² − 30x + 36 = 6(x² − 5x + 6).
Set f′(x) = 0: (x − 2)(x − 3) = 0, so x = 2 or x = 3.
f″(x) = 12x − 30.
At x = 2: f″ = 24 − 30 = −6, which is negative, so x = 2 gives a maximum.
At x = 3: f″ = 36 − 30 = 6, positive, so a minimum.
Maximum value f(2) = 2(8) − 15(4) + 36(2) + 10 = 16 − 60 + 72 + 10 = 38.

Answer: (a) 38

Example 2

The total cost of producing x units is C(x) = 2x² + 8x + 72. Find the output x that minimises the average cost. Options: (a) 4 (b) 6 (c) 9 (d) 12

Show the solution

Average cost AC = C(x) ÷ x = 2x + 8 + 72/x.
Differentiate: AC′ = 2 − 72/x².
Set AC′ = 0: x² = 36, so x = 6 (reject x = −6 since output cannot be negative).
AC″ = 144/x³. At x = 6, AC″ = 144/216, which is positive, so it is a minimum.

Answer: (b) 6

Example 3

A firm's revenue is R(x) = 50x − x² and its cost is C(x) = 10x + 100, where x is the number of units. The output that maximises profit is: (a) 10 (b) 15 (c) 20 (d) 25

Show the solution

Profit P(x) = R(x) − C(x) = 50x − x² − 10x − 100 = 40x − x² − 100.
P′(x) = 40 − 2x. Set it to zero: x = 20.
P″(x) = −2, which is negative, so x = 20 gives a maximum.
Check with marginal rule: R′ = 50 − 2x and C′ = 10. Equal when 50 − 2x = 10, so x = 20. This agrees.
Maximum profit is P(20) = 800 − 400 − 100 = 300, if the question asks for it.

Answer: (c) 20

Exam tips

Read the last line first. Check if the question asks for x or for the value of the function. Examiners put both quantities in the options.
For quadratics, use x = −b ÷ 2a and save the second derivative step.
In profit questions, the shortcut 'marginal revenue equals marginal cost' gives x quickly. Still check that P″ is negative.
Always reject negative or zero output when the context is production.
With 0.25 negative marking, guess only after you remove at least one or two options by finding the critical point.

Practice questions from Differential and Integral Calculus

Maxima and Minima: frequently asked questions

What is the difference between local maxima and minima?

A local maximum is a point where the function is higher than at nearby points. A local minimum is where it is lower than at nearby points. They describe only the neighbourhood, not the whole graph.

How do I find the maximum and minimum value using differentiation?

Find f′(x) and solve f′(x) = 0 to get critical points. Use f″(x) to classify each point as a maximum (negative) or minimum (positive). Then substitute the x-value into f(x) to get the value.

What if the second derivative is zero at the critical point?

The second derivative test fails. Use the first derivative test: check whether f′ changes sign from plus to minus (maximum) or minus to plus (minimum) around the point. If it does not change sign, it is neither.

How do I solve cost, revenue and profit problems?

Write profit as revenue minus cost, then maximise it using the derivative steps. For average cost, divide cost by x first and then minimise that function.