# Homework Help: MATLAB: Volume/Surface Area of Paper Cup

1. Apr 5, 2010

### malindenmoyer

A paper cup shpaed as a frustum of a cone with $$R_2=1.3R_1$$ is designed to have a volume of 240 cm3. Determine $$R_1$$ and height $$h$$of the cup such that the least amount of paper will be used for making the cup.

The problem must be done in MATLAB, so please keep that in mind. We have learned commands such as feval, fminbnd, syms, and function handles...all of which I think could be applicable to the problem.

Last edited: Apr 5, 2010
2. Apr 5, 2010

### Staff: Mentor

Show us the MATLAB code you have written. You won't get much help until you try something first.

3. Apr 5, 2010

### malindenmoyer

I am not particularly sure if this is even how you start the problem, but this is what I have done so far.

Code (Text):
syms R1 h;
R2=1.3*R1;
V=(pi*h)/3*(R1^2+R2^2+R1*R2)-240;
S=pi*(R1+R2)*sqrt((R2-R1)^2+h^2)+pi*R1^2;
I know that I could solve for $$h$$ using the Volume equation, and then have the Surface Area equation in terms of one variable essentially, as R2 is defined by R1; my thought process is such that I would take the new Surface Area equation (in terms of only R1) and take its derivative to find the critical points. This would give me R2 and then I could solve for h. Is this the right process?

Per LCKurtz's request, the two formulas I used to write the above code were the following:
$$V=\frac{1}{3}\pi h(\left R_1^2+R_2^2+R_1R_2)\right$$
$$S=\pi(R_1+R_2)\sqrt{(\left R_2-R_1)\right^2+h^2}+\pi R_1^2$$

Last edited: Apr 5, 2010
4. Apr 5, 2010

### LCKurtz

I don't use Matlab, but still...in your statement of the problem you give R2 = 1.3R1. Then in your code you put R2 = R1/3. Which is it? You need to be more careful.

For starters, your equation for volume needs to be just that, an equation, not a function.