Q. 225.0( 2 Votes )

# Prove that the vo

Answer :

Given,

• Volume of the sphere.

• Volume of the cone.

• Cone is inscribed in the sphere.

• Volume of cone is maximum.

Let us consider,

• The radius of the sphere be ‘a’ units.

• Volume of the inscribed cone be ‘V’.

• Height of the inscribed cone be ‘h’.

• Radius of the base of the cone is ‘r’.

Given volume of the inscribed cone is,

Consider OD = (AD-OA) =(h-a)

Now let OC^{2} = OD^{2} + DC^{2}, here OC = a, OD = (h-a), DC = r,

So a^{2} = (h-a)^{2} + r^{2}

r^{2} = a^{2} – (h^{2}+ a^{2} – 2ah)

r^{2} = h (2a -h) ----- (1)

Let us consider the volume of the cone:

Now substituting (1) in the volume formula,

---- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with h and then equating it to zero. __This is because if the function V(r) has a maximum/minimum at a point c then V’(c) = 0.__

Differentiating the equation (2) with respect to h:

[Since ]

------- (3)

To find the critical point, we need to equate equation (3) to zero.

4πah – 3πh^{2} = 0

h(4πa-3πh)= 0

h = 0 (or)

[as h cannot be zero]

__Now to check if this critical point will determine the maximum volume of the inscribed cone, we need to check with second differential which needs to be negative.__

Consider differentiating the equation (3) with h:

----- (4)

[Since ]

Now let us find the value of

As , so the function V is maximum at

Substituting h in equation (1)

As V is maximum, substituting h and r in the volume formula:

Therefore when the volume of a inscribed cone is maximum, then it is equal to times of the volume of the sphere in which it is inscribed.

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