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

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

## Comments

## Leave a comment