ISRO EC 2023 Q.95: solution and answer

ISRO 2023 · ECQ.95 · 2 marks · MCQ

What is the maximum area of the rectangle with perimeter 620 mm? (2 Marks)

(A)
24,025 mm^2
(B)
22,725 mm^2
(C)
24,000 mm^2
(D)
24,075 mm^2

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Answer and solution

ANSWER

The maximum area is $24, 025 mm^{2}$ , option (A).

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

Let the sides be $l$ and $b$ . The perimeter is $2 (l + b) = 620 mm$ , so $l + b = 310 mm$ .
The area is $A = l b = l (310 - l)$ .
Differentiate: $\frac{d A}{d l} = 310 - 2 l = 0$ , so $l = 155 mm$ . The second derivative is $- 2 < 0$ , so this is a maximum.
Then $b = 310 - 155 = 155 mm$ , so the rectangle is a square. (AM-GM gives the same result: $l b \leq (\frac{l + b}{2})^{2} = 15 5^{2}$ , with equality when $l = b$ .)
$A_{m a x} = 15 5^{2} = 24025 mm^{2}$ .
Option (a) equals this value. Options (b) and (c) are smaller than $24025$ , so they are not the maximum. Option (d), $24075$ , is larger than $24025$ , which no rectangle of this perimeter can reach.

KEY IDEA

For a fixed perimeter, the rectangle of largest area is a square: with $l + b$ fixed, $l b \leq (\frac{l + b}{2})^{2}$ .

WATCH OUT FOR

Using the full perimeter 620 as $l + b$ instead of half of it (310), or computing $(620/4)^{2}$ incorrectly. Another slip is picking a round number such as 24,000, which is an achievable area but not the largest.

From the ISRO EC 2023 paper. Also Q.95 of the ISRO Scientist/Engineer Computer Science 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 2 ISRO Geometry questions, with notes