Q. 295.0( 1 Vote )

# Show that the max

Answer :

Given,

• Radius of the sphere is .

• Volume of cylinder is maximum.

Let us consider,

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

• Volume of the inscribed cylinder be ‘V’.

• Height of the inscribed cylinder be ‘h’.

• Radius of the cylinder is ‘r’.

Now let AC^{2} = AB^{2} + BC^{2}, here AC = 2R, AB =2r, BC = h,

So 4R^{2} = 4r^{2} + h^{2}

----- (1)

Let us consider, the volume of the cylinder:

V = πr^{2}h

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(h) 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.

3h^{2}π = 4R^{2}π

h = 10

[as h cannot be negative]

__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 h=10

Substituting h in equation (1)

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

V = π (50) (10)

V = 500π cm^{3}

Therefore when the volume of a inscribed cylinder is maximum and is equal 500π cm^{3}

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