MathsCBSE Class 12 · Board exam 2026, set 65/5/1 · 4 marks · Case-based

CBSE Class 12 Mathematics 2026: question 38

Set in the CBSE Class 12 board exam of 2026, paper 65/5/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedApplication of Derivatives

A company produces cylindrical tumblers, open from the top. Since they want uniformity in the product, they fix the surface area of the tumblers produced.

Based on the above information, answer the following questions :

If for a tumbler, V is its volume, h the height and r the radius of the circular base, then :

(i)
Differentiate its volume with respect to radius of the base, where the surface area is constant.
2 marks
(ii)
If the company wants to maximize the volume of each tumbler, then establish a relation between its height and the radius of the base.
2 marks

Figure in the paper: A right circular cylinder labelled 'Tumbler', open at the top, with height h marked by a double-headed arrow on the right side and radius r marked on the circular base with centre O. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 65/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

An open-top cylinder has only one circular face: $S = 2 π r h + π r^{2}$ .

Hint 2

Eliminate h using S, so that V is a function of r alone, then differentiate.

Hint 3

Put $\frac{d V}{d r} = 0$ , confirm a maximum using the second derivative, and substitute S back into the expression for h.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) The tumbler is open at the top, so the surface area is curved surface plus base:
$S = 2 π r h + π r^{2}$ (constant). Hence $h = \frac{S - π r ^{2}}{2 π r}$ .
½ mark
(i) $V = π r^{2} h = π r^{2} \cdot \frac{S - π r ^{2}}{2 π r} = \frac{S r - π r ^{3}}{2}$ .
1 mark
(i) $\frac{d V}{d r} = \frac{S - 3 π r ^{2}}{2}$ .
½ mark
(ii) For a stationary point, $\frac{d V}{d r} = 0$ , which gives $S = 3 π r^{2}$ .
1 mark
(ii) $\frac{d ^{2} V}{d r ^{2}} = \frac{- 6 π r}{2} = - 3 π r < 0$ (since $r > 0$ ), so V is maximum when $S = 3 π r^{2}$ .
½ mark
(ii) Then $h = \frac{3 π r ^{2} - π r ^{2}}{2 π r} = \frac{2 π r ^{2}}{2 π r} = r$ .
So for maximum volume, the height equals the radius of the base: $h = r$ .
½ mark

Answer: (i) $\frac{d V}{d r} = \frac{S - 3 π r ^{2}}{2}$ ; (ii) $h = r$

Where the marks are

(i) Writes $S = 2 π r h + π r^{2}$ for the open cylinder (½ mark)
(i) Expresses V in terms of r: $V = \frac{S r - π r ^{3}}{2}$ (1 mark)
(i) Differentiates to get $\frac{d V}{d r} = \frac{S - 3 π r ^{2}}{2}$ (½ mark)
(ii) Puts $\frac{d V}{d r} = 0$ to get $S = 3 π r^{2}$ (1 mark)
(ii) Shows $\frac{d ^{2} V}{d r ^{2}} = - 3 π r < 0$ , so V is maximum (½ mark)
(ii) Concludes $h = r$ (½ mark)

Revise the chapter

Chapter: Application of Derivatives (notes, formulas, where marks are lost and every past question)
Topic: Maxima and Minima

Practise this chapterBoard-style questions from Application of Derivatives, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 65/5/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Mathematics 2026: question 38

Hints

Step-by-step solution

Revise the chapter

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