Question

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is $\dfrac{{2R}}{{\sqrt 3 }}$.​ Also find the maximum volume.

Answer 1

Hint : Here, the sphere is a shape that has radius and It is ball like structure. The cylinder is another shape that has two ends with circular shape and connected to a tube structure, so it has length, diameter of the circle. Now in this solution, find the maximum height that will be responsible for the maximum volume of the sphere, for that assume the diameter of the cylinder as x after that, substitute all values in the volume of the cylinder formula to find the maximum volume of the cylinder. So it can be inscribed in a sphere shape.

Complete step-by-step answer :
Given:
The radius of the sphere is $R = \dfrac{{2R}}{{\sqrt 3 }}$.
Let us assume the diameter of the cylinder is x.
Then we know that,

{x^2} + {h^2} = 4{R^2}\\\ {x^2} = 4{R^2} - {h^2} \end{array}$$ We know the equation to find the volume of the cylinder is $$V = \pi {r^2}h$$. As we know that the radius is half of the diameter, then volume of the cylinder can be written as, $$\begin{array}{c} V = \pi {\left( {\dfrac{x}{2}} \right)^2}h\\\ V = \dfrac{1}{4}\pi {x^2}h\\\ V = \dfrac{1}{4}\pi h\left( {4{R^2} - {h^2}} \right) \end{array}$$…….1 Differentiating the above equation with respect to the h, then, $$\begin{array}{l} \dfrac{{dV}}{{dh}} = \dfrac{d}{{dh}}\left( {\dfrac{1}{4}\pi h\left( {4{R^2} - {h^2}} \right)} \right)\\\ \dfrac{{dV}}{{dh}} = \pi {R^2} - \dfrac{\pi }{4}\left( {3{h^2}} \right)\\\ \dfrac{{dV}}{{dh}} = \pi \left( {{R^2} - \dfrac{3}{4}{h^2}} \right) \end{array}$$ Now equating the above equation, then we get, $$\begin{array}{c} \dfrac{{dV}}{{dh}} = 0\\\ \pi \left( {{R^2} - \dfrac{3}{4}{h^2}} \right) = 0\\\ h = \dfrac{{2R}}{{\sqrt 3 }} \end{array}$$ So, it is clear that the volume will be maximum at $$h = \dfrac{{2R}}{{\sqrt 3 }}$$. On substitute the values in the equation 1 to find the maximum volume, then we get, $$\begin{array}{c} = \dfrac{1}{4}\pi \dfrac{{2R}}{{\sqrt 3 }}\left( {4{R^2} - \dfrac{{4{R^2}}}{3}} \right)\\\ = \dfrac{{\pi R}}{{2\sqrt 3 }}\left( {\dfrac{{8{R^2}}}{3}} \right)\\\ = \dfrac{{4\pi {R^3}}}{{3\sqrt 3 }} \end{array}$$ Therefore, the maximum volume is $$V = \dfrac{{4\pi {R^3}}}{{3\sqrt 3 }}$$. **Note** : Make sure, while differentiating the values of the volume, after differentiation the h can be in the fraction that will be little difficult to find the maximum volume with the help of it, be careful while solving the problem, due to length procedure there may be chance of mistakes.