# Optimization problem using derivatives

1. Apr 8, 2013

### physicsernaw

1. The problem statement, all variables and given/known data

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.

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

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

3. 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?

2. Apr 8, 2013

### arildno

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

3. Apr 8, 2013

### physicsernaw

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

4. Apr 8, 2013

### 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.

5. Apr 8, 2013

### 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, noone can give you a "short cut", because we cannot see the algebraic pathways you actually chose to walk along..

6. Apr 8, 2013

7. Apr 8, 2013

### arildno

Well, lets 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.

8. Apr 8, 2013

### 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.

Last edited: Apr 8, 2013
9. Apr 8, 2013

### physicsernaw

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

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

10. Apr 8, 2013

### arildno

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