Calculus I, Section 4.7, #32
Optimization Problems
A right circular cylinder is inscribed in a cone with height h and base radius r. Find the largest possible
volume of such a cone.1
At right are four sketches of various cylinders inscribed a cone of height h and radius r. From
these sketches, it seems that the volume of the cylinder changes as a function of the cylinder’s radius,
x.
h
x
x
r
So we need a function that gives the volume of the
cylinder in terms of x.
Volumecylinder = (area of basecylinder ) heightcylinder
Volumecylinder = πx2 heightcylinder
h
r
h
h
x
x
r
r
In the diagram at right, we’ll let
y
y = height between top of cylinder and top of cone
h
h-y
h − y = height of the cylinder
Finally, we’ll use the diagram of the similar triangles
at right to find the height of the cylinder in terms of
x.
h
y
=
x
r
h
y
·x= ·x
x
r
h
y= x
r
x
r
y
h
x
r
If we substitute this value into our volume function, we get
Volumecylinder = πx2 heightcylinder
= πx2 (h − y)
h
2
h− x
= πx
r
3
πx h
= πx2 h −
r
so
V (x) = πx2 h −
πh 3
x
r
Now we compute the derivative, find the critical points and determine if the critical points give a local
maximum or local minimum.
1 Stewart,
Calculus, Early Transcendentals, p. 338, #32.
Continued =⇒
Calculus I
Optimization Problems
V (x) = πx2 h −
πh 3
x
r
so
3πh 2
V ′ (x) = 2πxh −
x
r 3
= πxh 2 − x
r
so
x=0
or x =
2
r
3
Clearly, x = 0 does not give a maximum volume, so we test 23 r
V ′′ (x) = 2πh −
6πh
x
r
and if x = 23 r,
V ′′
2
r
3
= 2πh −
6πh 2
· r
r
3
= 2πh − 4πh
= −2π
<0
so by the second derivative test, the critical value x = 32 r gives a maximum volume of
V
2
r
3
3
2
2
πh 2
=π
r h−
r
3
r
3
h 8 3
4 2
r h−π
r
=π
9
r 27
12
8
= π r2 h − π r2 h
27
27
4 2
=
πr h
27
Thus the maximum volume of a cylinder inscribed in a cone of radius r and height h is
4
2
27 πr h.