A regular square based pyramid is inscribed in a sphere of given radius `R` so that all vertices of the pyramid belong to the sphere. Find the greatest value of the volume of the pyramid.

Answer

`(64)/(81)r^(3)`

Explanation

If we cut the sphere from the center along the diagonal of the square base we get the view show in figure

Let r be the radius of the spere Then
Height of the pyramid =r+r `cos theta`
Diagonal of the square base =`2r sin theta`
`therefore` side of the squre base =`(2rsin theta)/sqrt(2)`
Let v be the volume of the pyaramid .Then
`V=1/3` (Area of the base xx Height)
`=1/3(2r sin theta)/(sqrt(2))^(2)xx(r+ r cos theta)`
`therefore (dV)/(d theta)=2/3r^(3)[sin 2 theta(1+cos theta) - sin ^(3)theta]`
`=2/3 r^(3)[2 sin 2 theta cos^(2)theta - sin ^(3)theta]`
`=2/3 r^(3)xx sin theta(2 cos theta + 2 cos^(2)theta - sin^(2)theta)`
`=2/3 r^(3)sin theta (3 cos^(2)theta +2 cos theta -1)`
`=2/3 r^(3)sin theta (cos theta +1)(3 cos theta -1)`
For maximum or minimum values of v we must have `(dV)/(d theta)=0`
or `3 cos theta - 1 =0`
cos `theta =1/3`
Hence V is maximum when cos `theta =1/3 and sin theta =(2sqrt(2))/(3)`
The maximum value of V is given by
`V=2/3r^(3)xx8/9(1+(1)/(3))=(64)/(81)r^(3)`