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Implicit differentiation and optimization

  1. Jul 3, 2010 #1
    1. The problem statement, all variables and given/known data

    A conical tent must contain [tex]40\pi ft^{3}[/tex]. Compute the height and radius of the tent with minimal total surface area. (Include the floor material.)

    2. Relevant equations
    1. [tex]\frac{\pi r^{2} h}{3} = 40\pi[/tex]
    2. [tex] \pi r \sqrt{r^{2} + h^{2}} + \pi r^{2} = S[/tex]
    3. [tex] \frac {dr}{dh} = - \frac {r}{2h} [/tex]
    4. [tex] \frac {dS} {dh} = \frac {2h^{2} - r^{2}}{2h\sqrt{r^{2} + h^{2}}} - \frac {r^{2}} {h}[/tex]

    3. The attempt at a solution
    I used implicit differentiation with respect to h in equation 1 to get equation 3. I then used implicit differentiation and substitution of equation 3 to get equation 4. Because I am looking for an extrema, I then set [tex]\frac {dS}{dh}[/tex] equal to zero and re-arrange to get:

    [tex] \frac {r^{2}} {h} = \frac {2h^{2} - r^{2}}{2h\sqrt{r^{2} + h^{2}}} [/tex]

    After this, though, I'm not sure how to manipulate equation 5 to get one variable in terms of the other. My book gives the answers [tex]h =4\sqrt [3]{15} ft[/tex], [tex]r = \sqrt{2}\sqrt[3]{15}[/tex], which could well be correct, but I'm unclear as to how the answer was derived.
  2. jcsd
  3. Jul 3, 2010 #2


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    Setting dS/dh= 0 is not enough. It has to be the "total differential" that is equal to 0. Yes,
    [tex]S(h,r)= \pi r \sqrt{r^{2} + h^{2}} + \pi r^{2}[/tex].

    And, since
    [tex]\frac{\pi r^{2} h}{3} = 40\pi[/tex]

    [itex]h= 120/r^2[/itex]. replace h in the surface area formula, so that S is a function of r only, differentiate that with respect to r, and set equal to 0.
  4. Jul 3, 2010 #3


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    In 4 you must remember that r is a function of h so you can't treat it as constant. For example, the derivative of r2 would be 2rdr/dh etc. You will need to substitute for your dr/dh from 3.
  5. Jul 3, 2010 #4
    OK, thanks. I think I understand now.
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