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

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October 6th, 2014, 07:54 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 06:47 PM. October 6th, 2014, 07:58 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 06:46 PM. October 7th, 2014, 02:30 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 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, 06:31 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 8th, 2014, 12:01 AM   #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? Tags confused, textbook Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post WWRtelescoping Complex Analysis 5 April 27th, 2014 01:50 PM layd33foxx Math Books 2 December 29th, 2011 07:57 PM badatmath Linear Algebra 6 July 13th, 2011 01:01 PM lemon5863 Math Books 1 February 8th, 2011 09:13 PM Kiranpreet Algebra 2 August 4th, 2008 02:10 PM

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