Show that the height of the cylinder of maximum volume, which can be inscribed in a sphere of radius R is `(2R)/sqrt3.` Also find the maximum volume.
Solution
Given, radius of the sphere is R.
Let r and h be the radius and the height of the inscribed cylinder respectively.
From the given figure, we have `h=2sqrt(R^2-r^2)`
The volume (V) of the cylinder is given by,
`V=pir^2h=2pir^2sqrt(R^2-r^2)`
`therefore (dV)/(dr)=4pirsqrt(R^2-r^2)+(2pir^2(-2r))/(2sqrt(R^2-r^2))`
`=(4pirR^2-6pir^3)/sqrt(R^2-r^2)`
for maxima or minima, `(dV)/(dr) =0 `
`(4pirR^2-6pir^3)=0`
`r^2=(2R^2)/3`
Now `,(d^2V)/(dr^2)=(sqrt(R^2-r^2)(4piR^2-18pir^2)-(4piR^2-6pir^3)(-2r)/(2sqrt(R^2-r^2)))/(R^2-r^2)`
`=((R^2-r^2)(4piR^2-18pir^2)-r(4piR^2-6pir^3))/(R^2-r^2)^(3/2)`
`=(4piR^4-22pir^2R^2+12pir^4+4pir^2R^2)/(R^2-r^2)^(3/2)`
Now, it can be observed that when `r^2=(2RR62)/3,(d^2V)/(dr^2)<0`
∴The volume is the maximum when `r^2=(2R^2)/3`
Maximum volume = V = `pih((4R^2 - h^2)/4)`
`h = 2R/sqrt3`
`V_(max) = pi xx 2R/sqrt3 ((4R^2 - 4R^2/3)/4)`
` = (2piR)/sqrt3 . (2R^2)/3 = (4piR^3)/(3sqrt3)` cubic units
Hence, the volume of the cylinder is the maximum when the height of the cylinder is `(2R)/3`