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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,

VolumeArea=(area of the base)(height)=r2h=(area of the base)+(area of the cylindrical surface)=r2+2rh.

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:

Minimise the area A(r)=r2+r2V, when r0.

The area function A(r) 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

A(r)=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,

A(r)=2+r34V

has the value

A3V=2+4VV=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

rh=3Vand=Vr2=VV−23=V1−23=V13=3V.