Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4r/3 .
Answers
Step-by-step explanation:
ANSWER
Let R and h be the radius and height of the cone.
r be the radius of sphere.
To show h=
3
4r
and Maximum volume of sphere =
27
8
Volume of sphere
In ΔABC, AC = h - r
Therefore, using Pythagoras theorem,
(h−r)
2
+R
2
=r
2
⇒R
2
=r
2
−(h−r)
2
Volume of cone: V =
3
1
πR
2
h
or V=
3
1
π(r
2
−(h−r)
2
)h
V=
3
1
π[r
2
−h
2
−r
2
+2hr]h
V=
3
1
π[2h
2
r−h
3
]
For maxima or minima,
dh
dV
=0
Now,
dh
dV
=
3
1
π[4hr−3h
2
]
Putting,
dh
dV
=0
we get
4hr=3h
2
h=
3
4r
dh
2
d
2
V
=
3
1
π[4r−6h]
Putting h=
3
4r
dh
2
d
2
V
=
3
1
π[4r−
3
6.4r
]⇒−
3
1
π[4r]
Which is less than zero, therefore h=
3
4r
is a maxima and the volume of the cone at h=
3
4r
will be maximum,
V=
3
1
πR
2
h
=
3
1
π[r
2
−(h−r)
2
]h
=
3
1
π[r
2
−(
3
4r
−r)
2
][
3
4r
]
=
3
1
π[
9
8r
2
][
3
4r
]
=
27
8
(
3
4πr
3
)
=
27
8
xvolume of sphere