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Minimizing Surface Area

  • Thread starter Sheneron
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  • #1
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[SOLVED] Minimizing Surface Area

1. Homework Statement
A can is to be manufactured in the shape of a circular cylinder with volume = 50.
Find the dimensions of a can that would minimize the amount of material needed to make the can.


2. Homework Equations
V = [tex] \pi r^2 h[/tex]
SA = [tex] 2 \pi r^2 + 2 \pi r h [/tex]


3. The Attempt at a Solution
I have never done a problem like this so I am unsure how to do it, but here is my attempt.

With the volume equation I solved for h. [tex] h = \frac{50}{\pi r^2} [/tex]
I plugged this value for h into the Surface area equation. [tex] SA = 2 \pi r^2 + 2 \pi r \frac{50}{\pi r^2} [/tex]

which = [tex] 2 \pi r^2 + \frac{100}{r} [/tex]

I then took the derivative of that and set it equal to 0.
[tex] 0 = 4 \pi r - 100 r^-2 [/tex]
[tex] r = \sqrt[3]{\frac{100}{4 \pi}} [/tex]
r = 1.996

Then I plugged that back into the volume equation to solve for h and got h= 3.99.
Could someone tell if this is right and if not where I went wrong. Thanks.
 

Answers and Replies

  • #2
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Looks good to me. Check your result by plugging in a number smaller than r and a number bigger than r into the derivative. If a number smaller than r makes the derivative negative and a number larger than r makes the derivative positive, then r would be a minimum.
 
  • #3
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So is that how I know that its a minimum instead of a maximum. I'm still a little confused about that. What would I do if I wanted to solve this problem for a maximum?
 
  • #4
HallsofIvy
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There is no maximum.
 
  • #5
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Does it not have one? Even though the volume is set to 50 there is no max Surface area?
 
  • #6
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If you keep increasing r, the surface area just gets bigger and bigger without bound. The height gets smaller, but there is a separate [tex]2\pi r^2[/tex] term in the surface area calculation. That's why there is no maximum surface area.
 
  • #7
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I understand I think. h could get infinitely small which would make the surface area infinitely large.
 

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