# Drinking cup

1. Apr 20, 2007

### Winzer

1. The problem statement, all variables and given/known data
A cone-shape drinking cup is made from a circular piece of paper of radius R by cutting out a sector and joining the edges CA and CB. Find the maximum capacityof such a cup. This is page#312 In James stewarts Calc book, 3rd edition by the way.

2. Relevant equations
$$V=\frac{1}{3}\pi r^2 h$$

3. The attempt at a solution
So realize this will form a cone with an inner triangle of hypotnue R. The base will have a radius r and a hieght h. So I will have to make a relationship between r and h to get R to plug into $$V=\frac{1}{3}\pi h r^2$$ and differeniate?

Last edited: Apr 20, 2007
2. Apr 20, 2007

### Dick

(1/3)*pi*r^2 doesn't even have the units of a volume. (1/3)*pi*r^2*h does. I think that's what you want. You will want to parametrize r and h by the angle cut out of R. And then differentiate and maximize wrt to the angle.

3. Apr 20, 2007

### Winzer

sorry clumsy mistake, I forgot the H, lol.

4. Apr 20, 2007

### Dick

I thought so. But just write r and h as functions of the cut out angle. That's just geometry.

5. Apr 21, 2007

### HallsofIvy

Staff Emeritus
Notice that the entire circle of paper has circumference $2\pi R$ but the circle at the top of the cone has circumference $2\pi r$. The arc length of the cut out wedge is $R \theta/\2\pi$ where $\theta$ is the angle in radians. That must be the difference between $2\pi R$ and $2\pi r$.

6. Apr 21, 2007

### Winzer

mmmm.......

7. Apr 22, 2007

### Dick

Ok, then let's try it another way. Draw the right triangle joining a point on the circle of the base of the cone to the apex of the cone. Height is h and base is r, right? What's the length of the hypotenuse? Use Pythagoras to eliminate one of the variables (r or h) from the formula for the volume and maximize.