Let H = height of the cone , R = radius of the base of the cone and X = distance from the centre of the sphere to to the base of the cone.
We can write the following equation:
R( exp2) + X (exp2) = 6 (exp2) (1)
The volume of the cone is given by the equation:
V = 1/3 PI R(exp2) H (2)
then, we have to maximize : R(exp2) H
From the equation (1), R(exp2) = 36 - X(exp2)
In addition, H = 6 + X
Therefore we obtain the function:
F(X) = (6 + X) ( 36 - X(exp2) or F(x) = 216 - 6 X(exp2) + 36 X - X(exp3)
The derivative of the function , d(F(X) /d X = -12 X + 36 - 3 X(exp2) This is a quadratic equation
At the maximum value for, a given X, the derivative equals cero
Then we have to solve the quadratic equation which yields two values X = 2 , X = -6
We take the value 2,
The cone of maximum volume has a height of 8 and a radius of sqrt ( 32)
By replacing values into the equation 2 , we obtain V = 268
