A rectangular visiting card is to contain 24 sq.cm. of printed matter. The margins at the top and bottom of the card are to be 1 cm and the margins on the left and right are to be 11/2 cm as shown below : On the basis of the above information, answer the following questions : (i) Write the expression for the area of the visiting card in terms of x. (ii) Obtain the dimensions of the card of minimum area.

(i) A(x)=2x+72/x+30; (ii) card is 9 cm × 6 cm

MathsCBSE Class 12 · Board exam 2024, set 65/5/1 · 4 marks · Case-based

CBSE Class 12 Mathematics 2024: question 37

Set in the CBSE Class 12 board exam of 2024, paper 65/5/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedApplication of Derivatives

A rectangular visiting card is to contain 24 sq.cm. of printed matter. The margins at the top and bottom of the card are to be 1 cm and the margins on the left and right are to be $1 \frac{1}{2}$ cm as shown below :

On the basis of the above information, answer the following questions :

(i)
Write the expression for the area of the visiting card in terms of $x$ .
2 marks
(ii)
Obtain the dimensions of the card of minimum area.
2 marks

Figure in the paper: Two rectangles. Left: visiting card with handwritten text inside a dashed rectangle, margins 1 cm at top and bottom, 1½ cm at left and right. Right: same card with the shaded inner rectangle labelled 'Printed matter', width x (top) and height y (right), with the same margins marked. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 65/5/1.

Hints

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

Hint 1

Add the margins to the printed matter: the card's length is x + 3 and its breadth is y + 2.

Hint 2

Use $x y = 24$ to remove y and get the area in terms of x alone.

Hint 3

Set $A^{'} (x) = 0$ and confirm a minimum with the second derivative.

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 printed area is $x y = 24$ , so $y = \frac{24}{x}$ .
The card measures $(x + 3)$ cm by $(y + 2)$ cm, so its area is
$A (x) = (x + 3) (y + 2) = x y + 2 x + 3 y + 6 = 24 + 2 x + \frac{72}{x} + 6$
$\Rightarrow A (x) = 2 x + \frac{72}{x} + 30$ .
2 marks
(ii) Differentiate: $A^{'} (x) = 2 - \frac{72}{x ^{2}}$ and $A^{''} (x) = \frac{144}{x ^{3}}$ .
1 mark
Set $A^{'} (x) = 0$ : $x^{2} = 36$ , so $x = 6$ (since $x > 0$ ). This is the critical point.
½ mark
Check: $A^{''} (6) = \frac{144}{216} > 0$ , so the area is minimum at $x = 6$ , which gives $y = \frac{24}{6} = 4$ .
Dimensions of the card: length $= 6 + 3 = 9$ cm and breadth $= 4 + 2 = 6$ cm (minimum area $54$ cm²).
½ mark

Answer: (i) $A (x) = 2 x + \frac{72}{x} + 30$ ; (ii) card is 9 cm × 6 cm

Where the marks are

Uses $x y = 24$ and forms $A (x) = 2 x + \frac{72}{x} + 30$ (2 marks)
Finds $A^{'} (x)$ and $A^{''} (x)$ (1 mark)
Critical point $x = 6$ (½ mark)
$A^{''} (6) > 0$ , dimensions 9 cm × 6 cm (½ mark)

Revise the chapter

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

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 paper2024 board paper, set 65/5/1: every question, by section.

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CBSE Class 12 Mathematics 2024: question 37

Hints

Step-by-step solution

Revise the chapter

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