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Aus Online Mathematik Brückenkurs 2

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

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

=r2h

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

=r2+2h

The problem can then be formulated as: minimise the can's area, A=r2+2h, whilst at the same time keeping the volume, V=r2h, constant.

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

h=Vr2

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

A=r2+2rVr2=r2+r2V

The minimisation problem is then:

to minimise the area Ar=r2+r2V  A(r)=..., when r0.

The area function Ar  is differentiable for all r0 and the region of definition r0 has no endpoints ( r=0 does not satisfy r0 ), so the function can only assume extreme values at critical points.

The derivative is given by

Ar=2r−r22V ,

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

2r−r22V=02r=r22Vr3=Vr=3V 

For this value of r, the second derivative,

Ar=2+r34V , has the value

A3V=2+V4V=60 ,

which shows that r=3V  is a local minimum.

Because the region of definition, r0, is open (the endpoint r=0 is not included) and unlimited, we cannot directly say that the area is least when r=3V ; it could be the case that area becomes smaller when r0 or r. In this case, however, the area increases without bound as r0 or r, so r=3V  really is a global minimum.

The metal can has the least area for a given volume 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 }}