Question

Solve the following : Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is `(2"R")/sqrt(3)`. Also, find the maximum volume.

Answer 1

Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm.

Then from the figure,

`"R"^2 + (h/2)^2` = r²

∴ R²  = `r^2 - h^2/(4)`   ...(1)

Let V be the volume of the cylinder.
Then V = πR²h

= `pi(r^2 - h^2/(4))h`   ...[By (1)]

= `pi(r^2 - h^3/(4))`

∴ `"dV"/"dh" = pid/"dh"(r^2h - h^3/(4))`

= `pi(r^2 xx 1 - 1/4 xx 3h^2)`

= `pi(r^2 - 3/4h^2)`
and
`(d^2V)/("dh"^2) = pid/"dh"(r^2 - 3/4h^2)`

= `pi(0 - 3/4 xx 2h)`

= `-(3)/(2)pih`

Now, `"dV"/"dh" = 0  "gives", pi(r^2 - 3/4h^2)` = 0

∴ `r^2 - 3/4h^2` = 0

∴ `(3)/(4)h^2` = r² 

∴ h² = `(4r^2)/(3)`

∴ h = `(2r)/sqrt(3)`          ...[∵ h > 0]
and
`((d^2V)/(dh^2))_("at"  h = (2r)/sqrt(3)`

= `-(3)/(2)pi xx (2r)/sqrt(3) < 0`

∴ V is maximum at h = `(2r)/sqrt(3)`

If h = `(2r)/sqrt(3)`, then from (1)

R² = `r^2 - (1)/(4) xx (4r^2)/(3) = (2r^2)/(3)`

∴ volumeof the largest cylinder

= `pi xx (2r^2)/(3) xx (2r)/sqrt(3) = (4pir^3)/(3sqrt(3)`cu cm.

Hence, the volume of the largest cylinder inscribed in a sphere of radius 'r' cm = `(4R^3)/(3sqrt(3)`cu cm.