Maths Differentiation and Applications of Derivatives JEE (Advanced) / IIT - JEE Problems ( Previous Years ) Single Correct MCQ
Published on: August 12, 2026

A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. The lengths of the sides of the rectangular sheet are

A
24
B
32
C
45
D
60

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Text Solution

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The correct answer is:
CHECK THE SOLUTION.

(a,c)

Let λ = 8x, b = 15 x

∴ Volume = (8x – 2a) (15x – 2a) = 4a 3 – 46a 2 x + 120 ax 2

= 6a 2 – 46ax + 60 x 2

= 0

∴ x = 3 and

= 6a – 23x

< 0,

So, at x = 3 gives maxima

> 0

So, at x = gives minima.

= 0 when a = 5 given ( ∴ 4a 2 = 100 given for maximum volume)

at a = 5

by = 0

⇒ 6x 2 – 23x + 15 = 0

x = 3 or 5/6

So by x = 3 (for max volume)

8x = 24, 15x = 45

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