MathsCBSE Class 12 · Application of Derivatives

Maxima and Minima: Derivative Tests and Optimisation Problems (Class 12)

Updated 15 October 2026 · Fact-checked

Solve f′(x) = 0 to find critical points. At such a point c, f″(c) < 0 gives a local maximum and f″(c) > 0 a local minimum; if f″(c) = 0, check the sign change of f′. On a closed interval, compare f at the critical points and the end points.

At the peak f′ changes + to − and f″<0 (maximum); at the valley f′ changes − to + and f″>0 (minimum); both tangents are flat.

Illustration drawn for QuestionBuddy.

Understand Maxima and Minima

A function has a local maximum at $c$ if $f (c) \geq f (x)$ for all $x$ near $c$ , and a local minimum if $f (c) \leq f (x)$ near $c$ . The absolute (greatest or least) value is the largest or smallest over the whole domain. At a local extreme point of a differentiable function, the tangent is horizontal, so $f^{'} (c) = 0$ ; such points (together with points where $f^{'}$ does not exist) are critical points. Not every critical point is an extremum: $x^{3}$ has $f^{'} (0) = 0$ but no maximum or minimum there.

First derivative test (motivated by the graph). At a critical point $c$ :

$f^{'}$ changes from $+$ to $-$ : local maximum (the graph rises then falls).
$f^{'}$ changes from $-$ to $+$ : local minimum.
$f^{'}$ does not change sign: neither (a point of inflection, like $x^{3}$ at 0).

Second derivative test (quicker when $f^{''}$ is easy). If $f^{'} (c) = 0$ :

$f^{''} (c) < 0$ : local maximum (the graph bends down).
$f^{''} (c) > 0$ : local minimum (the graph bends up).
$f^{''} (c) = 0$ : the test fails; go back to the first derivative test.

**Absolute extrema on a closed interval $[a, b]$ .** A continuous function on a closed interval always attains an absolute maximum and minimum. Find the critical points inside $(a, b)$ , evaluate $f$ at them and at the ends $a$ and $b$ , and pick the largest and smallest values. No derivative test is needed, but you must not forget the end points.

Optimisation (word problems). These are the 4- and 5-mark questions:

1. Draw a figure and name the variables.
2. Write the quantity to be maximised or minimised (volume, area, cost, profit) as a formula.
3. Use the given condition (fixed perimeter, fixed volume, fixed sum) to write it in one variable.
4. Differentiate, set equal to zero, solve, and reject values outside the practical range.
5. Confirm the nature with the second derivative (or first derivative test), and answer the actual question (dimensions, the maximum value, or both).

It is often easier to optimise a simpler quantity with the same extreme point, such as $A^{2}$ instead of $A$ , or to use a parameter such as an angle.

The syllabus says the first derivative test is 'motivated geometrically' and the second derivative test is 'given as a provable tool'. This is NCERT Section 6.4 and the highest-scoring part of the chapter.

Maxima and Minima: formulas to remember

Necessary condition: $f has a local extremum at c, f differentiable \Rightarrow f^{'} (c) = 0$; The converse is false ( $x^{3}$ at 0).
First derivative test: $f^{'} : + \to - \Rightarrow max; f^{'} : - \to + \Rightarrow min$; No sign change: neither.
Second derivative test: $f^{'} (c) = 0, f^{''} (c) < 0 \Rightarrow max; f^{''} (c) > 0 \Rightarrow min$; Inconclusive if $f^{''} (c) = 0$ .
Absolute extrema on [a, b]: $max / min of {f (a), f (b), f (c_{1}), f (c_{2}), \dots}$; $c_{i}$ are the critical points in $(a, b)$ .
Useful shapes: $V_{cyl} = π r^{2} h, S_{cyl} = 2 π r^{2} + 2 π r h, V_{box} = x (l - 2 x) (b - 2 x)$; Common in optimisation problems.
AM-GM check: $x + \frac{1}{x} \geq 2 (x > 0)$; A quick way to verify simple minima.

How to solve Maxima and Minima questions

Use this routine for local extrema, absolute extrema and optimisation problems.

1Find $f^{'} (x)$ , factorise, and solve $f^{'} (x) = 0$ for the critical points (keep only those in the domain or practical range).
2Local extrema: find $f^{''} (x)$ and evaluate it at each critical point; negative means maximum, positive means minimum. If zero, use the sign change of $f^{'}$ .
3Absolute extrema on $[a, b]$ : evaluate $f$ at every critical point inside the interval and at $a$ and $b$ ; compare the values.
4Word problems: express the target quantity in one variable using the constraint, then follow the steps above.
5Always state the final answer in words: the point, the maximum or minimum value, and the dimensions if asked.

Maxima and Minima in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Maximum area with fixed perimeter

Suppose a farmer has 120 m of fencing to enclose a rectangular pen for goats on open land.

Find the dimensions that give the largest area, and the largest area.

Try it first, then show the solutionSolution

Let the sides be $x$ and $y$ : $2 x + 2 y = 120$ , so $y = 60 - x$ .
Area $A = x (60 - x) = 60 x - x^{2}$ .
$A^{'} (x) = 60 - 2 x = 0 \Rightarrow x = 30$ ; $A^{''} (x) = - 2 < 0$ , so this is a maximum.
$y = 30$ ; $A = 900$ m².

Answer: A square of side 30 m, with area 900 m².

Memory hook: Fixed fence, biggest pen: make it a square.

Minimum material for a fixed volume

Imagine a company designing a closed cylindrical can that must hold $250 π$ cm³ of juice, using as little sheet metal as possible.

Find the radius and height that minimise the total surface area, and the least area.

Try it first, then show the solutionSolution

$π r^{2} h = 250 π \Rightarrow h = \frac{250}{r ^{2}}$ .
$S = 2 π r^{2} + 2 π r h = 2 π r^{2} + \frac{500 π}{r}$ .
$S^{'} (r) = 4 π r - \frac{500 π}{r ^{2}} = 0 \Rightarrow r^{3} = 125 \Rightarrow r = 5$ ; $S^{''} (r) = 4 π + \frac{1000 π}{r ^{3}} > 0$ , a minimum.
$h = \frac{250}{25} = 10$ cm; $S = 50 π + 100 π = 150 π$ cm².

Answer: $r = 5$ cm, $h = 10$ cm (height equals diameter); least area $150 π \approx 471$ cm².

Memory hook: The cheapest closed can is as tall as it is wide.

Tricks to remember Maxima and Minima

Short memory aids. Say each one aloud once, then use it on a question the same day.

Hill and valley
Up then down ( $+$ to $-$ ) is a hilltop: maximum. Down then up ( $-$ to $+$ ) is a valley: minimum.
Use it when: First derivative test.
Frown for max, smile for min
$f^{''} < 0$ : the curve frowns (maximum). $f^{''} > 0$ : it smiles (minimum).
Use it when: Second derivative test.
Ends are candidates too
On a closed interval, the end points can win. Always evaluate f at a and b.
Use it when: Absolute extrema questions.
One variable before differentiating
Use the constraint to eliminate the extra letter before you differentiate.
Use it when: Every optimisation word problem.

Maxima and Minima: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Local extrema of a cubic · 2 marks · Very short answerBoard level

Find the local maximum and local minimum values of $f (x) = x^{3} - 3 x + 2$ .

Try it first, then show the solutionSolution

$f^{'} (x) = 3 x^{2} - 3 = 0 \Rightarrow x = \pm 1$ ; $f^{''} (x) = 6 x$ . (1 mark)
$f^{''} (- 1) = - 6 < 0$ : local maximum $f (- 1) = - 1 + 3 + 2 = 4$ . $f^{''} (1) = 6 > 0$ : local minimum $f (1) = 1 - 3 + 2 = 0$ . (1 mark)

Answer: Local maximum 4 at $x = - 1$ ; local minimum 0 at $x = 1$ .

Example 2: Absolute extrema with ties · 3 marks · Short answerBoard level

Find the absolute maximum and minimum values of $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ on $[0, 4]$ .

Try it first, then show the solutionSolution

$f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ ; critical points $x = 1, 3$ , both in $[0, 4]$ . (1 mark)
$f (0) = 1$ , $f (1) = 5$ , $f (3) = 27 - 54 + 27 + 1 = 1$ , $f (4) = 64 - 96 + 36 + 1 = 5$ . (1 mark)
Absolute maximum 5 (at $x = 1$ and $x = 4$ ); absolute minimum 1 (at $x = 0$ and $x = 3$ ). (1 mark)

Answer: Absolute maximum 5 at $x = 1, 4$ ; absolute minimum 1 at $x = 0, 3$ .

Example 3: An open box from a rectangular sheet · 5 marks · Long answerHarder board level

An open box is made from a rectangular sheet 24 cm by 15 cm by cutting equal squares of side $x$ from each corner and folding up the sides. Find $x$ for maximum volume and the maximum volume.

Try it first, then show the solutionSolution

$V (x) = x (24 - 2 x) (15 - 2 x) = 4 x^{3} - 78 x^{2} + 360 x$ , with $0 < x < 7.5$ . (1 mark)
$V^{'} (x) = 12 x^{2} - 156 x + 360 = 12 (x^{2} - 13 x + 30) = 12 (x - 3) (x - 10)$ . (1 mark)
$V^{'} (x) = 0 \Rightarrow x = 3$ or $10$ ; reject $x = 10$ (outside $0 < x < 7.5$ ). (1 mark)
$V^{''} (x) = 24 x - 156$ ; $V^{''} (3) = - 84 < 0$ , so $x = 3$ gives a maximum. (1 mark)
$V (3) = 3 \times 18 \times 9 = 486$ cm³. (1 mark)

Answer: $x = 3$ cm; maximum volume 486 cm³.

Example 4: Maximising a product under a sum condition · 3 marks · Short answerHarder board level

Two positive numbers $x$ and $y$ have sum 30. Find them so that $x^{2} y$ is as large as possible.

Try it first, then show the solutionSolution

$y = 30 - x$ , so $P (x) = x^{2} (30 - x) = 30 x^{2} - x^{3}$ , $0 < x < 30$ . (1 mark)
$P^{'} (x) = 60 x - 3 x^{2} = 3 x (20 - x) = 0 \Rightarrow x = 20$ (since $x > 0$ ). (1 mark)
$P^{''} (x) = 60 - 6 x$ , $P^{''} (20) = - 60 < 0$ : maximum. $y = 10$ , $P = 400 \times 10 = 4000$ . (1 mark)

Answer: $x = 20$ , $y = 10$ ; maximum value 4000.

Example 5: Largest rectangle inside an ellipse · 5 marks · Long answerHOTS / competency

Find the largest area of a rectangle with sides parallel to the axes inscribed in the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1$ , and its dimensions.

Try it first, then show the solutionSolution

Let the corner in the first quadrant be $(4 cos θ, 3 sin θ)$ , $0 < θ < \frac{π}{2}$ ; this point lies on the ellipse. (1 mark)
Area $A = (2 \cdot 4 cos θ) (2 \cdot 3 sin θ) = 48 sin θ cos θ = 24 sin 2 θ$ . (1 mark)
$A^{'} (θ) = 48 cos 2 θ = 0 \Rightarrow θ = \frac{π}{4}$ . (1 mark)
$A^{''} (θ) = - 96 sin 2 θ$ ; at $\frac{π}{4}$ it is $- 96 < 0$ : maximum. (1 mark)
Maximum area $= 24$ ; sides $8 cos \frac{π}{4} = 42$ and $6 sin \frac{π}{4} = 32$ . (1 mark)

Answer: Maximum area 24 square units, with sides $42$ and $32$ .

Example 6: Case study: pricing concert tickets · 4 marks · Case-basedHOTS / competency

Suppose a school sells 500 tickets for a concert at Rs 200 each. A survey suggests that each Rs 10 reduction in price sells 50 more tickets. Let $n$ be the number of Rs 10 reductions.
(i) Write the revenue $R (n)$ . (1 mark)
(ii) Find $n$ for maximum revenue. (1 mark)
(iii) Confirm it is a maximum and find the best price and the maximum revenue. (2 marks)

Try it first, then show the solutionSolution

(i) $R (n) = (200 - 10 n) (500 + 50 n) = 100000 + 5000 n - 500 n^{2}$ . (1 mark)
(ii) $R^{'} (n) = 5000 - 1000 n = 0 \Rightarrow n = 5$ . (1 mark)
(iii) $R^{''} (n) = - 1000 < 0$ : maximum. Price $200 - 50 =$ Rs 150, tickets $500 + 250 = 750$ . (1 mark)
Maximum revenue $= 150 \times 750 =$ Rs 1,12,500. (1 mark)

Answer: (i) $R = 100000 + 5000 n - 500 n^{2}$ (ii) $n = 5$ (iii) price Rs 150, revenue Rs 1,12,500.

Where students lose marks in Maxima and Minima

The slips examiners see most often, and what to write instead.

Mistake: Forgetting the end points
On a closed interval, the absolute extremum may occur at an end point.
Wrong working
$f (x) = x^{3} - 6 x^{2} + 9 x + 1$ on $[0, 4]$ : values at critical points 5 and 1, so absolute max 5 only at $x = 1$ .
Right working
Also $f (4) = 5$ and $f (0) = 1$ : max 5 at $x = 1, 4$ ; min 1 at $x = 0, 3$ .
Do this: Make a table of f at critical points and both ends.
Usually costs: 1 to 2 marks.
Mistake: Not rejecting impossible critical values
Algebra may give values outside the practical range.
Wrong working
Box: $x = 3$ or $x = 10$ ; compare both.
Right working
$x = 10$ would need corners longer than half the 15 cm side, so reject it; $0 < x < 7.5$ .
Do this: State the domain of the variable before differentiating.
Usually costs: 1 mark.
Mistake: Skipping the nature test
A critical point may be a maximum, a minimum or neither.
Wrong working
$V^{'} (x) = 0$ at $x = 3$ , so maximum volume 486.
Right working
$V^{''} (3) = - 84 < 0$ , so $x = 3$ gives a maximum; maximum volume 486 cm³.
Do this: Always show $f^{''} (c)$ or the sign change of $f^{'}$ .
Usually costs: 1 mark in 5-mark questions.
Mistake: Keeping two variables
A function of two variables cannot be differentiated in one go here.
Wrong working
$A = x y$ , $\frac{d A}{d x} = y = 0$ .
Right working
Use the constraint, for example $y = 60 - x$ , so $A = 60 x - x^{2}$ .
Do this: Substitute the constraint first.
Usually costs: Most of the question.

Exam tips

Optimisation word problems appeared as 4- or 5-mark questions in every board paper from 2023 to 2026; draw a figure and state the domain of the variable.
Write a clear final sentence: 'Hence the volume is maximum when $x = 3$ cm, and the maximum volume is 486 cm³.'
Use the second derivative test when $f^{''}$ is simple; switch to the first derivative test if $f^{''} (c) = 0$ .
In absolute extrema questions, present the values in a small table; it makes comparison and marking easy.
Simplify before differentiating: maximising $A^{2}$ or using an angle parameter often saves time.

Practice questions on Maxima and Minima

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The minimum value of $x + \frac{1}{x}$ for $x > 0$ is:
(a) 1
(b) 2
(c) $\frac{1}{2}$
(d) 0
Hint
$f^{'} (x) = 1 - \frac{1}{x ^{2}}$ .
Answer
(b) 2, at $x = 1$ (where $f^{''} (1) = 2 > 0$ ).
Q2 · 2 marks · Very short answerBoard level
Find the maximum value of $f (x) = 3 sin x + 4 cos x$ .
Hint
$f^{'} (x) = 0$ gives $tan x = \frac{3}{4}$ ; take the angle in the first quadrant.
Answer
$f^{'} (x) = 3 cos x - 4 sin x = 0 \Rightarrow tan x = \frac{3}{4}$ . With $sin x = \frac{3}{5}$ , $cos x = \frac{4}{5}$ : $f = \frac{9}{5} + \frac{16}{5} = 5$ ; $f^{''} = - f = - 5 < 0$ , so the maximum value is 5.
Q3 · 2 marks · Very short answerBoard level
Find the local extreme value of $f (x) = x e^{x}$ .
Hint
$f^{'} (x) = e^{x} (1 + x)$ , $f^{''} (x) = e^{x} (2 + x)$ .
Answer
$f^{'} (x) = 0 \Rightarrow x = - 1$ ; $f^{''} (- 1) = e^{- 1} > 0$ , so a local minimum $f (- 1) = - \frac{1}{e}$ .
Q4 · 3 marks · Short answerHarder board level
Find the absolute maximum and minimum values of $f (x) = x - 2 sin x$ on $[0, 2 π]$ .
Hint
$f^{'} (x) = 1 - 2 cos x = 0$ at $\frac{π}{3}$ and $\frac{5 π}{3}$ .
Answer
$f (0) = 0$ , $f (\frac{π}{3}) = \frac{π}{3} - 3 \approx - 0.69$ , $f (\frac{5 π}{3}) = \frac{5 π}{3} + 3 \approx 6.97$ , $f (2 π) = 2 π \approx 6.28$ . Absolute maximum $\frac{5 π}{3} + 3$ ; absolute minimum $\frac{π}{3} - 3$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $f (x) = x^{3}$ has neither a maximum nor a minimum at $x = 0$ , although $f^{'} (0) = 0$ .
Reason (R): $f^{'} (x)$ does not change sign as x passes through 0.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
$f^{'} (x) = 3 x^{2} \geq 0$ on both sides of 0.
Answer
(a). $f^{'} (x) = 3 x^{2}$ is positive on both sides of 0, so by the first derivative test there is no extremum; R explains A.
Q6 · 5 marks · Long answerHarder board level
Find the height of the right circular cylinder of greatest volume that can be inscribed in a sphere of radius $33$ cm, and the greatest volume.
Hint
If the height is $h$ , the base radius satisfies $r^{2} = R^{2} - \frac{h ^{2}}{4}$ .
Answer
$V (h) = π (27 - \frac{h ^{2}}{4}) h = 27 π h - \frac{π h ^{3}}{4}$ . $V^{'} (h) = 27 π - \frac{3 π h ^{2}}{4} = 0 \Rightarrow h^{2} = 36 \Rightarrow h = 6$ cm. $V^{''} (h) = - \frac{3 π h}{2} < 0$ : maximum. $r^{2} = 27 - 9 = 18$ , so $V = π (18) (6) = 108 π$ cm³.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a gardener wants a rectangular vegetable bed along a straight river bank. No fence is needed along the river; the other three sides use 200 m of fencing.
(i) If each side perpendicular to the river is $w$ m, write the area $A (w)$ . (1 mark)
(ii) Find $w$ for maximum area. (2 marks)
(iii) Find the maximum area. (1 mark)
Hint
The side along the river is $200 - 2 w$ .
Answer
(i) $A (w) = w (200 - 2 w) = 200 w - 2 w^{2}$ . (ii) $A^{'} (w) = 200 - 4 w = 0 \Rightarrow w = 50$ m; $A^{''} = - 4 < 0$ , so maximum. (iii) Side along the river $= 100$ m; $A = 50 \times 100 = 5000$ m².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the necessary condition for a local extremum of a differentiable function at c?Show answer
$f^{'} (c) = 0$ .
2.If $f^{'} (c) = 0$ and $f^{''} (c) > 0$ , what does f have at c?Show answer
A local minimum.
3.What do you do if $f^{''} (c) = 0$ ?Show answer
Use the first derivative test (sign change of $f^{'}$ ).
4.Where can absolute extrema on $[a, b]$ occur?Show answer
At critical points inside the interval or at the end points.
5.For fixed perimeter, which rectangle has the largest area?Show answer
The square.
6.For a closed cylinder of fixed volume and least surface area, how are h and r related?Show answer
$h = 2 r$ .

Maxima and Minima: frequently asked questions

What is the difference between local and absolute maximum?

A local maximum is the highest value near a point; an absolute maximum is the highest value over the whole domain or interval. A function can have several local maxima but only one absolute maximum value.

When should I use the first derivative test instead of the second?

Use the first derivative test when the second derivative is hard to compute or equals zero at the critical point. It always works by checking whether f′ changes sign.

Do I need to check end points for local extrema?

No. End points matter for absolute extrema on a closed interval. Local extrema are found at critical points inside the domain.

Which NCERT section covers maxima and minima?

Section 6.4 (Maxima and Minima), including 6.4.1 on maximum and minimum values in a closed interval, in Chapter 6 of NCERT Mathematics Part I for Class 12.

How are optimisation problems marked?

Marks go to forming the function in one variable, differentiating and solving f′ = 0, confirming maximum or minimum, and stating the final answer with units.