What is the maximum volume of a cone inscribed in a sphere of radius 6?

Note: share your workings also

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What is the maximum volume of a cone inscribed in a sphere of radius 6?

Note: share your workings also

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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

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