Optimization problem using derivatives

[physicsernaw]

Homework Statement

We want to make a conical drinking cup out of paper. It should hold exactly 100 cubic inches of water. Find the dimensions of a cup of this type that minimizes the surface area.

Homework Equations

SA = pi*r^2 + pi*r*l where l is the slant height of the cone.

V = 1/3pi*r^2*h = 100

The Attempt at a Solution

I solve l in terms of h and r.

l = sqrt(r^2 + h^2)

Then I solve h in terms of r in the volume, and I get:

h = 300/pi*r^2

plugging this equation for h back in the equation for l, I get:

l = sqrt(r^2 + 300/(pi*r^2))

SA(r) = pi*r^2 + pi*r*sqrt(r^2 + 300/(pi*r^2))

Then taking the derivative and setting it equal to zero:

SA'(r) = 2pi*r + pi*sqrt(r^2 + 300/(pi*r^2)) + 1/2*pi*r/(sqrt(r^2 + 300/(pi*r^2)))*(2r + 600/(pi*r^3)) = 0

Instead of trying to solve that ugly thing myself I plugged it into wolfram and it said no real solutions exist so I'm obviously doing something wrong. Any help pleasE?

 

[arildno]

How can you drink from the cup when you have the lid on?

 

[physicsernaw]

Man, I hate math... lol. Thanks.

 

[physicsernaw]

Is there also an easier way to do this than the way I'm doing it? I get such long, ugly derivatives and finding the zeros is really tedious and time consuming... This is an exam review problem so I don't think the problem should take this long. Is there another approach I could use?

edit: I do get the right answer though, despite it taking a long time.

 

[arildno]

1. What you consider "long time" might not be felt like "long time" by others
2. Since you haven't really, posted the manner in which you've done your algebraic manipulations, no one can give you a "short cut", because we cannot see the algebraic pathways you actually chose to walk along..

 

[physicsernaw]

Here's what I got taking the derivative of the surface area, substituting h = 300/(pi*r^2)
and l = sqrt(r^2 + (300/(pi*r^2))^2)

SA(r) = pi*r*sqrt(r^2 + (300/(pi*r^2))^2)

dSA/dr = http://www.wolframalpha.com/input/?...)^2))*(2r+++600/(pi*r^2))*(-600/(pi*r^3))+=+0

 

[arildno]

Well, let's see if I can simplify the algebra a bit, on the spur of the moment:
$$\frac{dl}{dr}=\frac{r+h\frac{dh}{dr}}{l}$$
Thus, the optimizing equation becomes, effectively:
$$l+r\frac{dl}{dr}=0$$
That is:
$$l^{2}+r^{2}+rh\frac{dh}{dr}=0 (*)$$
Furthermore, we have:
$$\frac{dh}{dr}=-\frac{2}{r}h$$
therefore, (*) transmutes to:
$$2r^{2}-h^{2}=0$$

This ought to be correct simplfication.

 

[arildno]

Since $$h=\frac{3V}{\pi{r^{2}}}$$, we have:
$$r^{3}=\frac{3V}{\sqrt{2}\pi}$$
that is,
$$r=(\frac{3}{\sqrt{2}\pi})^{\frac{1}{3}}V^{\frac{1}{3}}$$
as our relation for minimized surface, given fixed volume of the open cone.

 

[physicsernaw]

I would have never thought of doing it in that way..

Thanks for the insight and taking the time to help me..

 

[arildno]

I would have never thought of doing it in that way..

Thanks for the insight and taking the time to help me..

Very often, retaining symbols for butt-ugly expressions will simplify the computation relative to substituting the symbol with original butt-ugliness.