Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius `R` is `(2R)/(sqrt(3))` .

Options

A. `(2r)/sqrt(3)`
B. `(r)/sqrt(3)`
C. 2r
D. `sqrt(3r)`

Answer

B

Explanation

Let be the height R be the radius and v be the volume of cylinder
In `triangle` OAB we have
`r^(2) =R^(2)+((h)/(2))^(2)` ltrbgt Clearly V=`pi R^(2) h` ltrbgt `rarr V(h)=pi (r^(2))-(h^(2))/(4)h`
`rarr v(h)=pi(r^(2))-(3h^(2))/(4)`
For maximum put v(h)=0
`rarr h= (2r)/sqrt(3)`
Differentiatin `ge` qu (ii) w.r.t h we get
`V''(h)=pi((-6h)/(4))`
``rarr V''(2r)/sqrt(3)=pi(-6)/(4)xx(2r)/sqrt(3)lt0`
Thus the volume is maximum when h =`(2r)sqrt(3)`