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Thread: Substition Within an Optimization Problem

  1. #1
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    Substition Within an Optimization Problem

    I am working on an optimization problem right now and ran into the following situation.

    $\displaystyle V = piR^2H = 8$
    $\displaystyle H = 8/piR^2$
    $\displaystyle H = 2srt(2)/piR$

    $\displaystyle A = H2piR+piR^2$
    $\displaystyle A = (2rt(2)/piR)2piR+piR^2$
    $\displaystyle A = 4srt(2)+piR^2$

    $\displaystyle dA/dR = 2piR$

    I think I have done this wrong, but I am fairly tired right now and can't seem to find my mistake.
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  2. #2
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    Quote Originally Posted by Vanilla View Post
    I am working on an optimization problem right now and ran into the following situation.

    $\displaystyle V = piR^2H = 8$
    $\displaystyle H = 8/piR^2$
    $\displaystyle H = 2srt(2)/piR$ incorrect

    $\displaystyle A = H2piR+piR^2$
    $\displaystyle A = (2rt(2)/piR)2piR+piR^2$
    $\displaystyle A = 4srt(2)+piR^2$

    $\displaystyle dA/dR = 2piR$

    I think I have done this wrong, but I am fairly tired right now and can't seem to find my mistake.
    Are you trying to minimise the surface area of the cylinder for a fixed volume of 8 cubic units?
    Last edited by Archie Meade; Oct 11th 2010 at 03:21 PM.
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  3. #3
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    Not sure where I got the idea that $\displaystyle piR^2 = (piR)^2$.

    Quote Originally Posted by Archie Meade View Post
    Are you trying to maximise the surface area of the cylinder for a fixed volume of 8 cubic units?
    Trying to minimize the surface area. However; I know how to solve the rest of the problem. Thanks for the help.

    Edit: Actually, I got stuck again.

    $\displaystyle A = H2piR + piR^2$ $\displaystyle H = 8/piR^2$
    $\displaystyle A = (8/piR^2)*2piR + piR^2$
    $\displaystyle A = (16piR/piR^2) + piR^2$
    $\displaystyle A = (16/R) + piR^2$
    $\displaystyle dA/dR = (R - 16/R^2) + 2piR$

    Then solving for dA/dR = 0.

    $\displaystyle 0 = (R - 16/R^2) + 2piR$
    $\displaystyle -2piR = (R - 16)/R^2$
    $\displaystyle -2piR^3 = R - 16$

    And from there I can't find a way to isolate R, probably because I made some silly algebraic error.
    Last edited by Vanilla; Oct 11th 2010 at 03:04 PM.
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  4. #4
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    Hi Vanilla,

    The volume of a cylinder is $\displaystyle {\pi}R^2H$

    The curved surface area is $\displaystyle 2{\pi}RH$

    The flat surface area is $\displaystyle {\pi}R^2$

    The total surface area of a cylinder closed at both ends (surface area of a solid cylinder) is

    $\displaystyle 2{\pi}RH+2{\pi}R^2$

    $\displaystyle V={\pi}R^2H=8\Rightarrow\ H=\displaystyle\frac{8}{{\pi}R^2}$

    Substitute the equation for H into A to get an equation for SA in terms of the single variable R...

    Surface area $\displaystyle =\displaystyle\ 2{\pi}R\ \frac{8}{{\pi}R^2}+2{\pi}R^2=\frac{16}{R}+2{\pi}R^ 2$

    Differentiating with respect to R and setting derivative to zero
    finds the value of R that gives minimum surface area.

    Double check whether or not the cylinder is closed.
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  5. #5
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    The cylinder is open on one end. So the $\displaystyle 2{\pi}R^2$ in the surface area equation would instead be $\displaystyle {\pi}R^2$, correct?
    So differentiating $\displaystyle \frac{16}{R}+{\pi}R^2$ would still leave me with $\displaystyle \frac{R-16}{R^2}+2{\pi}R$ for which I can't seem to isolate R when solving for 0.
    Last edited by Vanilla; Oct 11th 2010 at 04:13 PM.
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  6. #6
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    Yes,

    you only need to adjust for that before calculating the derivative.
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  7. #7
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    Double post.
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  8. #8
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    I think I got it.
    $\displaystyle \frac{16}{R}+{\pi}R^2$ can be simpified to $\displaystyle \frac{16+{\pi}R^3}{R}$, which then differentiates to $\displaystyle \frac{2{\pi}R^3-16}{R^2}$. And for $\displaystyle \frac{dA}{dR} = 0$, $\displaystyle R = (\frac{8}{\pi})^\frac{1}{3}$. Thanks for the help!
    Last edited by Vanilla; Oct 11th 2010 at 04:37 PM.
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  9. #9
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    Quote Originally Posted by Vanilla View Post
    I think I got it.
    $\displaystyle \frac{16}{R}+{\pi}R^2$ can be simpified to $\displaystyle \frac{16+{\pi}R^3}{R}$, which then differentiates to $\displaystyle \frac{2{\pi}R^3-16}{R^2}$. And for $\displaystyle \frac{dA}{dR} = 0$, $\displaystyle R = (\frac{8}{\pi})^\frac{1}{2}$. Thanks for the help!
    Yes,

    but as the numerator is zero for R giving minimum area, you need to take the cube root
    rather than the square root.
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