Aus Online Mathematik Brückenkurs 2

If we call the radius of the metal can \displaystyle r and its height \displaystyle h, then we can determine the can's volume and area by using the figures below.

Volume = (area of the base). (height)

\displaystyle =\pi r^{2}h

Area= (area of the base)+(area of the cylindrical surface)

\displaystyle =\pi r^{2}+2\pi h

The problem can then be formulated as: minimise the can's area, \displaystyle A=\pi r^{2}+2\pi h, whilst at the same time keeping the volume, \displaystyle V=\pi r^{2}h, constant.

From the formula for the volume, we can make \displaystyle h the subject,

\displaystyle h=\frac{V}{\pi r^{2}}

and express the area solely in terms of the radius, \displaystyle r:

\displaystyle A=\pi r^{2}+2\pi r\bullet \frac{V}{\pi r^{2}}=\pi r^{2}+\frac{2V}{r}

The minimisation problem is then:

to minimise the area \displaystyle A\left( r \right)=\pi r^{2}+\frac{2V}{r} A(r)=..., when \displaystyle r>0.

The area function \displaystyle A\left( r \right) is differentiable for all \displaystyle r>0 and the region of definition \displaystyle r>0 has no endpoints ( \displaystyle r=0\text{ } does not satisfy \displaystyle r>0 ), so the function can only assume extreme values at critical points.

The derivative is given by

\displaystyle {A}'\left( r \right)=2\pi r-\frac{2V}{r^{2}},

and if we set the derivative equal to zero, so as to obtain the critical points, we get

\displaystyle \begin{align} & 2\pi r-\frac{2V}{r^{2}}=0\quad \Leftrightarrow \quad 2\pi r=\frac{2V}{r^{2}} \\ & \Leftrightarrow \quad r^{3}=\frac{V}{\pi }\quad \Leftrightarrow \quad r=\sqrt[3]{\frac{V}{\pi }} \\ \end{align}

For this value of \displaystyle r, the second derivative,

\displaystyle {A}''\left( r \right)=2\pi +\frac{4V}{r^{3}}, has the value

\displaystyle {A}''\left( \sqrt[3]{\frac{V}{\pi }} \right)=2\pi +\frac{4V}{\frac{V}{\pi }}=6\pi >0,

which shows that \displaystyle r=\sqrt[3]{{V}/{\pi }\;} is a local minimum.

Because the region of definition, \displaystyle r>0, is open (the endpoint \displaystyle r=0\text{ } is not included) and unlimited, we cannot directly say that the area is least when \displaystyle r=\sqrt[3]{{V}/{\pi }\;}; it could be the case that area becomes smaller when \displaystyle r\to 0 or \displaystyle r\to \infty . In this case, however, the area increases without bound as \displaystyle r\to 0 or \displaystyle r\to \infty , so \displaystyle r=\sqrt[3]{{V}/{\pi }\;} really is a global minimum.

The metal can has the least area for a given volume \displaystyle V when

\displaystyle r=\sqrt[3]{{V}/{\pi }\;}, and

\displaystyle h=\frac{V}{\pi r^{2}}=\frac{V}{\pi }\left( \frac{V}{\pi } \right)^{-{2}/{3}\;}=\left( \frac{V}{\pi } \right)^{1-{2}/{3}\;}=\left( \frac{V}{\pi } \right)^{{1}/{3}\;}=\sqrt[3]{\frac{V}{\pi }}