Math, asked by raju3701, 8 months ago

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

Answered by shravani7894
1

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

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