A right circular cylinder that is open at the top and has a given surface area will have the greatest volume if its height \[h\]and radius \[r\] are related by

A. \[2h = r\]
B. \[h = 4r\]
C. \[h = 2r\]
D. \[h = r\]

Hint: Simplify the formula of the volume of the cylinder, by using the formula of the surface area of the cylinder. Then differentiate the formula of volume with respect to \[r\]. To calculate the greatest volume, equate the differential function to zero and reach to the required answer.
Formula used :
Volume of the cylinder: \[V = \pi {r^2}h\]
Surface area of the cylinder: \[S = 2\pi rh + 2\pi {r^2}\]
Complete step by step solution:
Given: The height of a cylinder is \[h\] and radius is \[r\].
The given cylinder is open at top.
So, the surface area of the cylinder is,
\[S = 2\pi rh + \pi {r^2}\]
Solve the above equation and calculate the value of height \[h\].
\[h = \dfrac{{S - \pi {r^2}}}{{2\pi r}}\]
Now substitute the above value in the formula of volume of a cylinder.
\[V = \pi {r^2}\left( {\dfrac{{S - \pi {r^2}}}{{2\pi r}}} \right)\]
\[ \Rightarrow \]\[V = \left( {\dfrac{r}{2}} \right)\left( {S - \pi {r^2}} \right)\]
\[ \Rightarrow \]\[V = \dfrac{1}{2}\left( {Sr - \pi {r^3}} \right)\]
Now differentiate the above formula with respect to \[r\].
\[\dfrac{{dV}}{{dr}} = \dfrac{1}{2}\left( {S - 3\pi {r^2}} \right)\] \[.....\left( 1 \right)\]
Now to find the critical point, equate the first derivative to zero.
 \[\dfrac{{dV}}{{dx}} = 0\].
\[ \Rightarrow \]\[\dfrac{1}{2}\left( {S - 3\pi {r^2}} \right) = 0\]
\[ \Rightarrow \]\[S - 3\pi {r^2} = 0\] [ Since \[\dfrac{1}{2} > 0\] ]
\[ \Rightarrow \]\[S = 3\pi {r^2}\]
Now substitute \[S = 2\pi rh + \pi {r^2}\] in above equation.
\[2\pi rh + \pi {r^2} = 3\pi {r^2}\]
\[ \Rightarrow \]\[2\pi rh = 2\pi {r^2}\]
Divide both sides by \[2\pi r\].
\[h = r\]
To find the maximum or minimum substitute the critical point \[h = r\] in the second derivative.

Differentiate the equation \[\left( 1 \right)\] with respect to \[r\].
\[\dfrac{{{d^2}V}}{{d{r^2}}} = \dfrac{1}{2}\dfrac{d}{{dr}}\left( {S - 3\pi {r^2}} \right)\]
\[ \Rightarrow \]\[\dfrac{{{d^2}V}}{{d{r^2}}} = \dfrac{1}{2}\left( {0 - 6\pi r} \right)\]
\[ \Rightarrow \]\[\dfrac{{{d^2}V}}{{d{r^2}}} = - 3\pi r\]
Now substitute \[r = h\] in the above equation.
\[\dfrac{{{d^2}V}}{{d{r^2}}} = - 3\pi h\]
Since \[ - 3\pi h < 0\].
Hence, the cylinder has greatest volume when \[h = r\].
Hence the correct option is D.
Note: The maximum value of a function is the place where the function reaches its highest point.
A function \[f\left( x \right)\] has a maximum value at \[x = a\], if \[f'\left( a \right) = 0\], and \[f''\left( a \right) < 0\].
A function \[f\left( x \right)\] has a minimum value at \[x = a\], if \[f'\left( a \right) = 0\], and \[f''\left( a \right) > 0\].