Question

Given that the solid obtained by rotating a rectangle about one of its sides is a cylinder. If the perimeter of a rectangle is 48 cm and the volume of the cylinder formed by rotating it is maximum, then the dimensions of that rectangle is

• A. \( 14, 10 \)
• B. \( 20, 4 \)
• C. \( 18, 6 \)
• D. \( 8, 16 \)

Hint:

Let the sides of the rectangle be \( x \) and \( y \). Given \( 2(x + y) = 48 \implies x + y = 24 \). When rotated about side \( y \), volume \( V = \pi x^2 y = \pi x^2 (24 - x) \). Maximize \( V \) using calculus. Similarly, consider rotation about side \( x \). The dimensions that maximize the volume will be the same in both scenarios.

Answer 1

The Correct Option is D

Let the dimensions of the rectangle be length \( l \) and width \( w \).
The perimeter of the rectangle is given as 48 cm, so \( 2(l + w) = 48 \), which implies \( l + w = 24 \).
Case 1: Rotation about the width \( w \).
The radius of the cylinder formed is \( r = l \) and the height is \( h = w \).
The volume of the cylinder is \( V_1 = \pi r^2 h = \pi l^2 w \).
From \( l + w = 24 \), we have \( w = 24 - l \).
Substituting this into the volume equation: \( V_1(l) = \pi l^2 (24 - l) = 24\pi l^2 - \pi l^3 \).
To maximize the volume, we find the critical points by taking the derivative with respect to \( l \) and setting it to zero: \( \frac{dV_1}{dl} = 48\pi l - 3\pi l^2 = 3\pi l (16 - l) \).
Setting \( \frac{dV_1}{dl} = 0 \), we get \( l = 0 \) or \( l = 16 \).
Since \( l \) must be positive, \( l = 16 \).
The second derivative is \( \frac{d^2V_1}{dl^2} = 48\pi - 6\pi l \).
At \( l = 16 \), \( \frac{d^2V_1}{dl^2} = 48\pi - 6\pi (16) = 48\pi - 96\pi = -48\pi<0 \), so the volume is maximum at \( l = 16 \).
If \( l = 16 \), then \( w = 24 - 16 = 8 \).
The dimensions are \( 16, 8 \).
Case 2: Rotation about the length \( l \).
The radius of the cylinder formed is \( r = w \) and the height is \( h = l \).
The volume of the cylinder is \( V_2 = \pi r^2 h = \pi w^2 l \).
From \( l + w = 24 \), we have \( l = 24 - w \).
Substituting this into the volume equation: \( V_2(w) = \pi w^2 (24 - w) = 24\pi w^2 - \pi w^3 \).
Taking the derivative with respect to \( w \): \( \frac{dV_2}{dw} = 48\pi w - 3\pi w^2 = 3\pi w (16 - w) \).
Setting \( \frac{dV_2}{dw} = 0 \), we get \( w = 0 \) or \( w = 16 \).
Since \( w \) must be positive, \( w = 16 \).
If \( w = 16 \), then \( l = 24 - 16 = 8 \).
The dimensions are \( 8, 16 \).
In both cases, the dimensions that maximize the volume are 16 cm and 8 cm.