[SOLVED] Applications of Differentiation Question

**looi76** · April 28th 2008, 01:11 AM

Question:
A hollow circular cylinder, open at one end, is constructed of thin sheet metal. The total external surface area of the cylinder is $192\pi$ $cm^2$ . The cylinder has a radius of $r$ $cm$ and a height of $h$ $cm$ .

(i) Express h in terms of r and show that the volume, $V$ $cm^3$ , of the cylinder is given by:
$V = \frac{1}{2}\pi(192r - r^3)$

Given that $r$ can vary,
(ii) Find the value of $r$ for which $V$ has a stationary value,
(iii) find this stationary value and determine whether it is a maximum or a minimum.

(i) Can someone please show me how to get $V = \frac{1}{2}\pi(192r - r^3)$ ?

**Danshader** · April 28th 2008, 01:20 AM

external surface area of a closed cylinder
= $2\pi r h + 2\pi r^2$

external surface area for one open end of cylinder
= $2\pi r h + \pi r^2$
= $192\pi$

obtain an expression for h.

volume of cylinder
= $\pi r^2 h$

substitute the value of h into the volume equation

**looi76** · April 28th 2008, 02:04 AM

Originally Posted by Danshader

external surface area of a closed cylinder
= $2\pi r h + 2\pi r^2$

external surface area for one open end of cylinder
= $2\pi r h + \pi r^2$
= $192\pi$

obtain an expression for h.

volume of cylinder
= $\pi r^2 h$

substitute the value of h into the volume equation

Thanks Danshader

$2\pi{rh} + \pi{r^2} = 192\pi$

$\frac{2\pi{rh}}{\pi} + \frac{\pi{r^2}}{\pi} = \frac{192\pi}{\pi}$

$2rh + r^2 = 192$

$2rh = 192 - r^2$

$h = \frac{192 - r^3}{2r}$

$= \pi{r^2h}$

$=\pi{r^2}\frac{192 - r^3}{2r}$

$V = \frac{1}{2}\pi{}(192r - r^3)$

Now how do I find the value of r?

**Isomorphism** · April 28th 2008, 02:12 AM

Definition:A point 'u' at which the derivative of a function f(x) vanishes,i.e. f'(u) = 0 is called a stationary point.

This means you have to find values of r such that V'(r) = 0.

**Danshader** · April 28th 2008, 02:26 AM

yup just like what isomorphism said, a stationary point can be found when the derivative of V with respect to r is 0

$\frac{dV}{dr} = 0$

earlier you already prove the equation of the volume:

$V = \frac{1}{2}\pi{}(192r - r^3)$

hence all you need to do is to differentiate this equation with respect to r and equate it to 0:

$\frac{d}{dr}{(\frac{1}{2}\pi{}(192r - r^3)}) = 0$

solve for r. good luck

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