optimization headache

ok so i have a cone with a surface area of 2 but thats all i know, so know i have to find what the maximum volume i can have with that cone, so before i take the derivitive and set it to 0 I need either radius in terms of height or height in terms of radius and i know surface area = pi*r*sqrt(r^2+h^2) but im having a hell of a time getting one in terms of the other, any help?

Oh, is it already time for optimization problems in high-school calculus?

Do your own homework.

>>2
actaully im trying to build a drinking cup for water but i only have 2m^2 of material and im thirsty so i wanna do this right so i dont have to keep getting up as much to refill my cup

Surface area on the inside or outside? what is the distance between the inside and outside of the cone surfaces?

>>2
Down here in Aus it's the end of the year.

>>1
2 = πr²√(r²+h²)

1. Divide both sides by πr²
2. Square both sides
3. Get h² in terms of r
4. Root both sides, take the positive as h must be positive. You now have h in terms of r
5. V = ¹/₃πr²h
6. Substitute h from 4 into 5
7. Diffrentiate to get ^dV/_dr. Use the product and chain rules or a calulator.
8. Let ^dV/_dr = 0
9. Solve for r or graph it and get the r intercepts. Take the positive as r must be positive.
10. Substitute r back into the original eqation to gt h.
DONE

It's a bit long but easily doable.

>>5
Oh it's only:
2 = πr√(r²+h²)

That makes things easier. You only need to use the chain rule when differentiating now.