Question

A rectangle of a perimeter of 24 inches is rotated about one of its sides so as to form a cylinder. To attain the maximum volume of the cylinder what would be the dimensions of the rectangle (in inches).

Answer 1

Hint: Given problem tests the concepts of derivatives and their applications. These types of questions can be easily solved if we keep in mind the concepts of maxima and minima. By the formula of the perimeter of the rectangle and the volume of the cylinder, we obtain the volume of the cylinder in terms of a single variable and then use the maxima and minima concepts to find the maximum volume of the cylinder. 

Formula used:
Perimeter of rectangle $= 2\left( {length + breadth} \right)$
Volume of cylinder $V = \pi {r^2}h\,\,c{m^3}$

Complete step by step Solution:
Given, A rectangle of the perimeter is 24 inches.
Let $x$ be the length and $y$ be the breadth of the rectangle or dimension of the rectangle.
Perimeter of rectangle $= 2\left( {x + y} \right)$
$24 = 2\left( {x + y} \right)$
$x + y = \dfrac{{24}}{2}$
$x + y = 12$
$y = 12 - x$ ----- (1)
We know that
Volume of cylinder $V = \pi {r^2}h\,$
If we make a cylinder from a rectangle. The breadth of the rectangle becomes the height of the cylinder and length becomes radius. i.e., $y = h$ and $x = r$ then,
$V = \pi {x^2}\left( {12 - x\,} \right)$
$V = 12\pi {x^2} - \pi {x^3}$

Differentiating with respect to x on both sides to find the critical points where the maximum and minimum values occur.
$\dfrac{{dV}}{{dx}} = 24\pi x - 3\pi {x^2}$
Finding critical points by equating $\dfrac{{dV}}{{dx}} = 24\pi x - 3\pi {x^2}$ to zero.
$24\pi x - 3\pi {x^2} = 0$
$\Rightarrow \pi \left( 24x-3{{x}^{2}} \right)=0$
$\Rightarrow 3x\left( 8-x \right)=0$
Either $3x = 0$ or $8 - x = 0$
Either $x = 0$ or $x = 8$
Length can’t be zero. So, $x = 8$.
For the second derivative, we differentiate $\dfrac{{dV}}{{dx}} = 24\pi x - 3\pi {x^2}$ again by x. So, we get,
$\dfrac{{{d}^{2}}V}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 24\pi x-3\pi {{x}^{2}} \right)=24\pi -6\pi x$
Substituting the value of x at critical point obtained from first derivative test, we get,
 $\dfrac{{{d}^{2}}V}{d{{x}^{2}}}=24\pi -6\pi \left( 8 \right)=-24\pi $
So, $x = 8$ is a point of maxima.
Substitute x in equation (1)
 $y = 12 - 8$
$y = 4$
Hence, the dimensions of the rectangle (in inches) are $x = 8$ and $y = 4$ or length$= 8$ inches and breadth $= 4$ inches.

Note:
To solve these kinds of problems we have to know about the formulas. In measurement, we have to remember to write a unit. For the perimeter, the unit will be the same as the unit of the given length and for the volume, the unit will be cubic units. Students must remember the units and they have to mention them. The first derivative test is used to find the critical points where the maxima and minima occur. The second derivative test is used to find which point corresponds to maxima and which points correspond to minima.