Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi-vertical angle α is one-third that of the come and the greatest volume of the cylinder is

OR

Find the intervals in which the function is strictly increasing or strictly decreasing.

Let VAB be the cone of height ‘h’ and semi-vertical angle α

Let x be the radius of the base of the cylinder A’B’CD which is inscribed in the right circular cone and H be the height

Then, Height of the cylinder, H = VO – VO’

H = (h – x cot α) …(i)

Now, Volume of a cylinder, V = πx²H

= πx²(h – x cot α)

= πx²h – πx³ cot α

Differentiating with respect to x, we get

…(ii)

For maxima or minima V,

⇒ 2πxh - 3πx² cot α = 0

⇒ 2πxh = 3πx² cot α

⇒ 2h = 3x cot α

Now, we again differentiate the eq. (ii), we get

When , we have

⇒ V is maximum when

Putting the value of x in eq. (i), we get

Height of the cylinder = of height of cone

Now, the maximum volume of the cylinder is:

V = πx²H

OR

Given:

x ∊ [0, 2π]

Step 1: First we find the f’(x)

Differentiate with respect to x, we get

[∵ cos²x + sin²x = 1]

Step 2: Putting f’(x) = 0

∴ cos x (4 – cos x) = 0

cos x = 0 or 4 – cosx = 0

x = cos^-1(0) or cos x = 4

or But -1 ≤ cos x ≤ 1

So, cos x = 4 is not possible

We know that,

Putting n=0

Putting n=1

Putting n=2

since, x ∊ (0, 2π)

So, value of x are

Step 3: plotting of value of x

Thus, we divide the interval (0, 2π) into three disjoint intervals

Step 4: now, we find the interval at which the given function is strictly increasing or strictly decreasing:

and