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Answer to Question #24805 in Calculus for Anita Horton

Question #24805
A right angled triangle of a given hypothernuse (h) is rotated about one of it's legs to generate a right circular cone. Show that the maximum volume is given by 2 sqrroot3 x pi x h^3 / 27
Expert's answer
Suppose,& right angled triangle has legs a and b:
a^2 + b^2 = h^2
and is rotated about a.
Therefore, the volume of a right circular cone equals:
V = 1/3 a*pi*b^2,
or (a=Sqrt[h^2-b^2])
V =(pi/3) b^2 Sqrt[[h^2-b^2]
we need find maximum, therefore dV/db = 0
dV/db = (pi/3) ( 2b*Sqrt[h^2-b^2] - b^2/Sqrt[h^2-b^2]*b ) = 0
2 (h^2 - b^2) = b^2 => b = Sqrt[2/3] h
a =& Sqrt[h^2-b^2] = Sqrt[1/3] h
V = 1/3 a*pi*b^2 = 1/3 pi h^3 2/3 /Sqrt[3] = 2*Sqrt[3]/27 pi h^3

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