My Math Forum Example in a Textbook, Which Got Me Confused

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October 6th, 2014, 06:54 AM   #1
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Example in a Textbook, Which Got Me Confused

Quote:
 A piece of aluminium will be made into a topless cylinder whose volume is $\displaystyle 8,000\pi\space cm^3$ Determine the cylinder's height and radius so that the aluminium will be used as few as possible. Answer: Given, the cylinder's volume = V(r), its height = h, its radius = r, and its surface area = S(r). V(r) = The area of the bottom $\displaystyle \times$ height $\displaystyle =\pi r^2\times h=8,000\pi$ so that $\displaystyle h=\frac{8,000\pi}{\pi r^2}=\frac{8,000}{r^2}$ S(r) = the area of the bottom + the area of the side = $\displaystyle \pi r^2+2\pi rh$ $\displaystyle S(r)=\pi r^2-2\pi r(\frac{8,000}{r^2})=\pi r^2-2\pi rh$ The stationary value of S(r) is obtained if S'(r) = 0 so $\displaystyle S'(r)=2\pi r-\frac{16,000\pi}{r^2}$
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Maybe I would just stop here because here's where I got lost. How come the denominator $\displaystyle r^2$ didn't get derived? Could we just take h as a constant and didn't take it into account when deriving in respect to r despite it has r within it? Shouldn't the derivative be $\displaystyle S'(r)=2\pi r-16,000\pi$ instead? Please give me some light on this.

Last edited by skipjack; October 7th, 2014 at 05:47 PM.

October 6th, 2014, 06:58 AM   #2
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Quote:
 Originally Posted by Monox D. I-Fly . Maybe I would just stop here because here's where I got lost. How come the denominator $\displaystyle r^2$ didn't get derived? Could we just take h as a constant and didn't take it into account when deriving in respect to r despite it has r within it? Shouldn't the derivative be $\displaystyle S'(r)=2\pi r-16,000\pi$ instead? Please give me some light on this.
h goes as 1/r^2. The last term in S(r) in terms of h goes as r. So when you plug h into the last term of S(r) it goes as 1/r. When you take the derivative the last term will then go as 1/r^2.

-Dan

Last edited by skipjack; October 7th, 2014 at 05:46 PM.

 October 7th, 2014, 01:30 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions Be careful with the sign; the derivative of $\displaystyle \frac{1}{r}$ with respect to $\displaystyle r$ is $\displaystyle -\frac{1}{r^2}$. Otherwise your calculation is fine.
October 7th, 2014, 05:31 PM   #4
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Quote:
 Originally Posted by Monox D. I-Fly $\displaystyle S(r)=\pi r^2-2\pi r(\frac{8,000}{r^2})=\pi r^2-2\pi rh$
That's incorrect.
$\displaystyle S(r)=\pi r^2+2\pi rh=\pi r^2+2\pi r(\frac{8,000}{r^2})=\pi r^2+\frac{16,000\pi}{r}$
(Obtaining S as a function of r is useful, as h is a function of r.)
You can proceed by use of differentiation or as shown below.
$\displaystyle S = \pi((r^3 - 1200r + 16,000)/r + 1,200) = \pi((r-20)^2(r + 40)/r + 1,200)$,
which is minimized (for $r$ > 0) when $r$ = 20.
The corresponding value of $h$ is also 20.
Hence the required radius is 20cm and the required height is 20cm.

October 7th, 2014, 11:01 PM   #5
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Quote:
 Originally Posted by skipjack That's incorrect. $\displaystyle S(r)=\pi r^2+2\pi rh=\pi r^2+2\pi r(\frac{8,000}{r^2})=\pi r^2+\frac{16,000\pi}{r}$ (Obtaining S as a function of r is useful, as h is a function of r.) You can proceed by use of differentiation or as shown below. $\displaystyle S = \pi((r^3 - 1200r + 16,000)/r + 1,200) = \pi((r-20)^2(r + 40)/r + 1,200)$, which is minimized (for $r$ > 0) when $r$ = 20. The corresponding value of $h$ is also 20. Hence the required radius is 20cm and the required height is 20cm.
Where did you get that -1200 from?

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