Question

A rectangular sheet of metal of length 6 meters and width 2 meters is given. Four equal square are removed from the corners. The sides of this sheet are now turned up to form an open rectangular box. The height of the box (in cm), such that the volume of the box is maximum is

• A. 22
• B. 25
• C. 35
• D. 45

Answer 1

The correct option is D 45


Let the side of each of the equare cut off be x m and the sides of the base are 6 - 2x, 2 - 2x m.

$\therefore$ Volume V of the box $=x(6-2x)(2-2x)$
$=4(x^3-4x^2+3x)$
Then, $\frac{dV}{dx}=4(3x^2-8x+3)$

For volume to be maximum

$\frac{dV}{dx}=0and\frac{d^{2}V}{d{x}^2}<0$
$3x^2-8x+3=0$

$\therefore$ $x=\frac{8\pm \sqrt{64-36}}{6}=\frac{8\pm \sqrt{28}}{6}=0.45mor2.2m$

2.2 m is not possible

For x = 0.45 m.

$\frac{d^{2}V}{d{x}^2}=4(6x-8)$

$=4(6\times 0.45-8)=-21.2<0$

Hence the volume of the box is maximum when its height is 45 cm.