Answer :

Clearly, from the figure, we need to prove that

r = R/2 ,where R is the radius of cone and r is the radius of a cylinder with maximum curved surface area inscribed inside the cone.

So in order to prove this, we need to find the curved surface area of the cylinder and maximise it using derivatives.

Clearly, from fig, we can say that-

The height of cylinder is = h

The height of cone = H

The radius of cone = R

Radius of cylinder = r

Let C represents curved surface area of the cylinder.

∴ C = 2πrh {applying formula for curved surface area of cylinder}

Here we have two variables h and r. To maximise it we need to express it only in terms of r as a result is desired for r.

Again observe the figure, we have 2 right triangles one with side R and other smaller one with side r.

Both share the same semi-vertical angle.

∴ tan θ =

∴ RH – Rh = rH

⇒ h =

∴ C =

For C to be maximum,

So, differentiating C w.r.t r, we get –

⇒ …(1)

∴

⇒ R – 2r = 0

⇒ R = 2r

Checking for maxima –

Differentiating equation 1 again-

⇒

∴ R = 2r gives the maxima.

Hence, the radius of cylinder r is half times the radius of cone R.

**OR**

Let h be the height and x be the side of the square base of the open box,

Total Area of box = x^{2}+ 4xh = C^{2} (given)

∴

We know that the volume of a cuboid is given by-

V = l × b × h

∴ V = x^{2}h

⇒ V () =

For volume to be maximum:

So, differentiating the above equation w.r.t x -

⇒

For maximum or minimum

⇒

⇒

∴ x = (Thus, V is max or min at x = )

∵ = -ve

∴ V is maximum at x =

Hence, h =

⇒ h = =

∴ V_{max} = x^{2}h = cubic units

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