Question

Show that the height of a cylinder open at the top of a given volume and minimum total surface area, is equal to the radius of the base.

Answer 1

Hint: To solve this question we will first use the formula of volume of cylinders and surface area of cylinder given as,
\[V=\pi {{r}^{2}}h\] & \[S=2\pi r\left( h+r \right)\], where h is height of cylinder and r is radius from volume V we will compute value of ‘h’ and substitute it in surface area S. Finally we will differentiate S and \[S'=\dfrac{dS}{dr}\] with respect to S to get the value of height h.

Complete step by step answer:
Let us first of all draw a cylinder of radius r and height h.
It is given as,

Let volume of cylinder be V.
The formula of volume of cylinder of height h and radius r is given by,
\[V=\pi {{r}^{2}}h\]
Let the surface area of the cylinder be S.
The formula of surface area of cylinder of radius r and height h is given by,
\[S=2\pi r\left( h+r \right)\]
Or \[S=2\pi rh+2\pi {{r}^{2}}\]
Now as our cylinder is or radius r and height ‘h’ then volume of cylinder as stated above is:
\[V=\pi {{r}^{2}}h\]
Dividing by \[\pi {{r}^{2}}\] we get:
\[\Rightarrow h=\dfrac{V}{\pi {{r}^{2}}}\]
\[\Rightarrow \] height of cylinder = \[h=\dfrac{V}{\pi {{r}^{2}}}\] - (1)
Now from formula of S (Surface area) as stated above we have,
\[S=2\pi rh+2\pi {{r}^{2}}\]
\[2\pi {{r}^{2}}\] is for both base and top. Now as we have open top,
\[\Rightarrow S=2\pi rh+\pi {{r}^{2}}\]
Substituting value of h as obtained in equation (1) we get,
\[\begin{align}
  & \Rightarrow S=2\pi r\left( \dfrac{V}{\pi {{r}^{2}}} \right)+\pi {{r}^{2}} \\
 & \Rightarrow S=\dfrac{2}{r}\left( V \right)+\pi {{r}^{2}} \\
\end{align}\]
\[\Rightarrow S=\dfrac{2V}{r}+\pi {{r}^{2}}\] - (2)
Now as we are given surface area S minimum then we will differentiate the above equation (2) with respect to radius r.
Differentiating with respect to r we get,
\[\begin{align}
  & \dfrac{dS}{dr}=\dfrac{d}{dr}\left( \dfrac{2V}{r} \right)+\left( \dfrac{d}{dr}\left( \pi {{r}^{2}} \right) \right) \\
 & \Rightarrow \dfrac{dS}{dr}=2V\left( \dfrac{-1}{{{r}^{2}}} \right)+2\pi r \\
\end{align}\]
\[\Rightarrow \dfrac{dS}{dr}=\dfrac{-2V}{{{r}^{2}}}+2\pi r\] - (3)
For the total area to be minimum, \[\dfrac{dS}{dr}=0\].
\[\begin{align}
  & \Rightarrow \dfrac{-2V}{{{r}^{2}}}+2\pi r=0 \\
 & \Rightarrow \dfrac{+2V}{{{r}^{2}}}=2\pi r \\
 & \Rightarrow \dfrac{V}{{{r}^{2}}}=\pi r \\
 & \Rightarrow V=\pi {{r}^{3}} \\
\end{align}\]
Now because the volume of the cylinder is \[V=\pi {{r}^{2}}h\].
Substituting this we get,
\[\begin{align}
  & \Rightarrow V=\pi {{r}^{3}} \\
 & \Rightarrow \pi {{r}^{2}}h=\pi {{r}^{3}} \\
\end{align}\]
Cancelling \[\pi {{r}^{2}}\] we have,
\[\Rightarrow h=r\] - (4)
Again differentiating equation (2) with respect to r we get,
\[\begin{align}
  & \dfrac{d}{dr}\left( \dfrac{dS}{dr} \right)=\dfrac{d}{dr}\left( \dfrac{-2V}{{{r}^{2}}} \right)+\dfrac{d}{dr}\left( 2\pi r \right) \\
 & \dfrac{{{d}^{2}}S}{d{{r}^{2}}}=-2V\left( \left( -2 \right){{r}^{-2-1}} \right)+2\pi \\
 & \Rightarrow \dfrac{{{d}^{2}}S}{d{{r}^{2}}}=\dfrac{4V}{{{r}^{3}}}+2\pi \\
\end{align}\]
Therefore, we have obtained that the height of the cylinder open at top of given volume and minimum total surface area is equal to the radius of base.

Note:
Whenever any value or expression is given to be minimum as here S was minimum then process to calculate is – Compute \[\dfrac{dS}{dr}\] and put \[\dfrac{dS}{dr}=0\] to get the value of height ‘h’ or radius ‘r’ as desired in question.
After that obtained \[\dfrac{{{d}^{2}}S}{d{{r}^{2}}}\] and observe is \[\dfrac{{{d}^{2}}S}{d{{r}^{2}}}>0\] then value of h obtained above is minimum and if \[\dfrac{{{d}^{2}}S}{d{{r}^{2}}}<0\] then value of h obtained is maximum.