calculus
posted by jj snoke .
A right triangle with hypotenuse 5 inches is revolved around one of its legs to generate a right circular cone. Find the radius and height of the greatest volume cone.

v = 1/3 pi r^2 h
r^2 + h^2 = 25
v = 1/3 pi (25h^2)h = 1/3 pi (25h  h^3)
a max/min occurs when dv/dh = 0:
25  3h^2 = 0
h = 5/√3
r = 5√2/√3
It's a max because v'' < 0
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