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?

>>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.