(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Implicit differentiation and optimization

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