Question

A right cylinder is inscribed in a sphere of radius “r”. Find the largest possible volume of such a cylinder.

Answer 1

Hint:In order to find the maximum volume of the cylinder which is to be inscribed in a sphere, the concept of maxima and minima is used. The function of volume of cylinder \[V = \pi {r^2}h\] in terms of height of sphere is to be differentiated twice first to get the critical points and then to check the sign of the double derivative.

Complete step by step solution:
We can illustrate the question in the following figure:
Note that “h” refers to half of the total height of the cylinder. We chose to use “h” instead
of \[\dfrac{h}{2}\] to simplify things later on.
To find the volume of our cylinder, we need to multiply the area of the top by the total height of the cylinder. In other words;
\[V = \pi {\left( {radius of cylinder} \right)^2}\left( {height of cylinder} \right)\]
Substituting the values we have,
\[
V = \pi {\left( {\sqrt {{r^2} - {h^2}} } \right)^2}\left( {2h} \right) \\
\Rightarrow V = 2\pi h\left( {{r^2} - {h^2}} \right) \\
\]
This is our volume function. Next we take the derivative of the volume function and set it equal to zero. If we move the h inside the parenthesis, we only need to use the power rule to get the derivative that is given below,
\[
V = 2\pi \left( {{r^2}h - {h^3}} \right) \\
\Rightarrow \dfrac{d}{{dx}}V\left( h \right) = 2\pi \left( {{r^2} - 3{h^2}} \right) = 0 \\
\]
The \[2\pi \]divides out and we are left with,
\[\left( {{r^2} - 3{h^2}} \right) = 0\]
After some rearranging we have,
\[h = \dfrac{r}{{\sqrt 3 }}\]
This is our optimized height. To find the optimized volume, we need to plug this into the volume function.
\[V = 2\pi h\left( {{r^2} - {h^2}} \right) = 2\pi \left( {\dfrac{r}{{\sqrt 3 }}} \right)\left( {{r^2} - {{\left( {\dfrac{r}{{\sqrt 3 }}} \right)}^2}} \right)\]
Simplifying it we have,
\[
\Rightarrow V = \dfrac{{2\pi r}}{{\sqrt 3 }}\left( {{r^2} - \dfrac{{{r^2}}}{3}} \right) \\
\Rightarrow V = \dfrac{{2\pi r}}{{\sqrt 3 }}\left( {\dfrac{{2{r^2}}}{3}} \right) \\
\Rightarrow V = \dfrac{{4\sqrt 3 \pi {r^3}}}{9} \\
\]
This is the optimized volume for the cylinder.

Note: To solve this problem the formula for volume of the cylinder should be remembered. Also the knowledge of Pythagoras theorem is required. To find the volume of our cylinder, we need to multiply the area of the top by the total height of the cylinder. The concept of maxima and minima is used in this problem.