Question

Diagram 5 shows a right cylinder with a diameter of 2xcm. Given that the total surface area of the cylinder is 96cm³.Find the maximum volume of the cylinder

Answer 1

Given:

The diameter of the right cylinder is 2x cm.

The total surface area is 96 cm cube.

The radius is calculated as,

$\begin{gathered} r=\frac{d}{2} \\ r=\frac{2x}{2} \\ r=x\text{ cm} \end{gathered}$

The total surface area is,

$\begin{gathered} S=2\pi rh+2\pi(r)^2 \\ 96=2\pi xh+2\pi(x^2) \\ h=\frac{96-2\pi(x^2)}{2\pi x} \end{gathered}$

Volume is,

$\begin{gathered} V=\pi(r)^2h \\ =\pi(x^2)\frac{96-2\pi(x^2)}{2\pi x} \\ =\frac{x(96-2\pi(x^2)}{2} \end{gathered}$

Now, differentiate with respect to x,

$\begin{gathered} \frac{dV}{dx}^{}=\frac{d}{dx}(\frac{x(96-2\pi(x^2)}{2}) \\ =\frac{d}{dx}\mleft(x\mleft(-\pi x^2+48\mright)\mright) \\ =\frac{d}{dx}\mleft(x\mright)\mleft(-\pi x^2+48\mright)+\frac{d}{dx}\mleft(-\pi x^2+48\mright)x \\ =1\cdot\mleft(-\pi x^2+48\mright)+\mleft(-2\pi x\mright)x \\ =84-3\pi(x^2)\ldots\ldots\ldots\ldots\text{.}(1) \end{gathered}$

Now,

$\begin{gathered} \frac{dV}{dx}=0 \\ 84-3\pi(x^2)=0 \\ x^2=\frac{16}{\pi} \\ x=\sqrt[]{\frac{16}{\pi}} \end{gathered}$

Now, differentiate (1) with respect to x again,

$\begin{gathered} \frac{d^2V}{dx^2}=\frac{d}{dx}(84-3\pi(x^2)) \\ =-6\pi x \\ At\text{ x=}\sqrt[]{\frac{16}{\pi}} \\ \frac{d^2V}{dx^2}=-6\pi\sqrt[]{\frac{16}{\pi}}<0 \\ \end{gathered}$

Since, the double derivative is negative.

$So,\text{ the volume is maximum at }\sqrt[]{\frac{16}{\pi}}$

So, the volume becomes,

$\begin{gathered} V=\pi(x^2)h \\ V=\pi(\sqrt[]{\frac{16}{\pi}})^2h \\ V=\frac{16h}{\pi} \end{gathered}$

Answer: maximum volume of the cylinder is,