JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Application of Derivatives

A box open from top is made from a rectangular sheet of dimension

a \times b

by cutting squares each of side

x

from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to:

Answer: (B)

Solution

After cutting squares of side

x

from each corner and folding up the flaps, the box has base

(a - 2 x) \times (b - 2 x)

and height

x

, so

V = x (a - 2 x) (b - 2 x) = 4 x^{3} - 2 (a + b) x^{2} + ab x .

Setting

\frac{d V}{d x} = 12 x^{2} - 4 (a + b) x + ab = 0

gives

x = \frac{4 ( a + b ) \pm 16 ( a + b ) ^{2} - 48 ab}{24} = \frac{( a + b ) \pm a ^{2} + b ^{2} - ab}{6} .

The larger root exceeds half the smaller side and gives a negative dimension, so the volume is maximised at the smaller root

x = \frac{a + b - a ^{2} + b ^{2} - ab}{6}

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option