Implicit differentiation and optimization

[Strants]

Homework Statement

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

Homework Equations

1. $$\frac{\pi r^{2} h}{3} = 40\pi$$
2. $$\pi r \sqrt{r^{2} + h^{2}} + \pi r^{2} = S$$
3. $$\frac {dr}{dh} = - \frac {r}{2h}$$
4. $$\frac {dS} {dh} = \frac {2h^{2} - r^{2}}{2h\sqrt{r^{2} + h^{2}}} - \frac {r^{2}} {h}$$

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 $$\frac {dS}{dh}$$ equal to zero and re-arrange to get:

$$\frac {r^{2}} {h} = \frac {2h^{2} - r^{2}}{2h\sqrt{r^{2} + h^{2}}}$$

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 $$h =4\sqrt [3]{15} ft$$, $$r = \sqrt{2}\sqrt[3]{15}$$, which could well be correct, but I'm unclear as to how the answer was derived.

 

[HallsofIvy]

Homework Statement

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

Homework Equations

1. $$\frac{\pi r^{2} h}{3} = 40\pi$$
2. $$\pi r \sqrt{r^{2} + h^{2}} + \pi r^{2} = S$$
3. $$\frac {dr}{dh} = - \frac {r}{2h}$$
4. $$\frac {dS} {dh} = \frac {2h^{2} - r^{2}}{2h\sqrt{r^{2} + h^{2}}} - \frac {r^{2}} {h}$$

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 $$\frac {dS}{dh}$$ equal to zero and re-arrange to get:

$$\frac {r^{2}} {h} = \frac {2h^{2} - r^{2}}{2h\sqrt{r^{2} + h^{2}}}$$

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 $$h =4\sqrt [3]{15} ft$$, $$r = \sqrt{2}\sqrt[3]{15}$$, which could well be correct, but I'm unclear as to how the answer was derived.

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

And, since
$$\frac{\pi r^{2} h}{3} = 40\pi$$

$h= 120/r^2$. 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.

 

[LCKurtz]

2. $$\pi r \sqrt{r^{2} + h^{2}} + \pi r^{2} = S$$
3. $$\frac {dr}{dh} = - \frac {r}{2h}$$
4. $$\frac {dS} {dh} = \frac {2h^{2} - r^{2}}{2h\sqrt{r^{2} + h^{2}}} - \frac {r^{2}} {h}$$

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.

 

[Strants]

OK, thanks. I think I understand now.