Lösung 1.3:6
Aus Online Mathematik Brückenkurs 2
If we call the radius of the metal can
Volume = (area of the base). (height)
r2h
Area= (area of the base)+(area of the cylindrical surface)
r2+2
h
The problem can then be formulated as: minimise the can's area,
r2+2
h
r2h
From the formula for the volume, we can make
r2
and express the area solely in terms of the radius,
r2+2
r
V
r2=
r2+r2V
The minimisation problem is then:
to minimise the area
r
=
r2+r2V
0
The area function
r
0
0
0
The derivative is given by
r
=2
r−r22V
and if we set the derivative equal to zero, so as to obtain the critical points, we get
r−r22V=0
2
r=r22V
r3=
V
r=
3
V
For this value of
r
=2
+r34V
3
V
=2
+
V4V=6
0
which shows that
3V
Because the region of definition,
0
3V
0
0
3V
The metal can has the least area for a given volume
3V
r2=
V
V
−2
3=
V
1−2
3=
V
1
3=
3
V