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Derivative problem

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A company wants to build a cylindrical can with a total surface area of 100 square inches, which includes the bottom and top of the can. Find the dimensions that maximize the volume.

    2. Relevant equations

    volume of a cylinder = 2*pi*r^2*h
    surface area of cylinder = 2*pi*r^2 + 2*pi*r*h


    3. The attempt at a solution

    I wrote a huge wall of text detailing how I solved the problem, but when I went to post I had gotten logged out due to how much time I spent on the post. So I'm going to make this short and just explicate my process:

    I took the formula for constant surface area 100 inches squared and solved for the height in terms of the radius. Then, I plugged that equation (h in terms of r) into the volume and took the derivative of the volume with respect to the radius.

    I set the derivative equal to zero to find the radius that gives the max volume, and I got
    r = sqrt(50/(pi*3))

    Then I plugged r back into the equation for h in terms of r, and simplifying, I got:
    h = sqrt(150/pi) - sqrt(50/(3*pi))

    -----------------------------------------------------------------------------------------------

    The answer sheet I have says that I should have gotten H = 2R. Indeed my answer reflects this, but should I have arrived at a general solution H = 2R instead of the exact answers I got? How would I go about solving this generally?

    Was my approach fine? Could someone maybe solve it for me in an alternate manner? I would really appreciate iany advice. Thank you.
     
  2. jcsd
  3. Apr 9, 2013 #2

    Simon Bridge

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    There a setting that stops you from being logged out like that ... though I used to write out my posts on a text editor and copy-and-paste it over ;)
    The question asks for the specific solution - the general relation would maximize the volume for any given surface area.
     
  4. Apr 9, 2013 #3

    HallsofIvy

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    If surface area [itex]S= 2\pi R H+ 2\pi R^2 [/itex] and you want to maximize the volume [itex]V= \pi R^2H[/itex] then [itex]\nabla V= 2\pi RH\vec{i}+ \pi R^2\vec{j}[/itex] must be a multiple of [itex]\nabla S= (2\pi H+ 4\pi R)\vec{i}+ 2\pi R\vec{j}[/itex]. That means we must have [itex]2\pi RH= \lambda(2\pi H+ 4\pi R)[/itex] and [itex]\pi R^2= 2\lambda\pi R[/itex] for some constant [itex]\lambda[/itex]. We can eliminate [itex]\lambda[/itex] by dividing the first equation by the second:
    [tex]\frac{2\pi RH}{\pi R^2}= \frac{2\pi H+ 4\pi R}{2\pi R}[/tex]
    [tex]\frac{2H}{R}= \frac{H+ 2R}{R}[/tex]
    [tex]2H= H+ 2R[/tex]
    [tex]H= 2R[/tex]
     
  5. Apr 9, 2013 #4
    Hey thanks!!!
     
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