Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^{- 1} √2

Solution:

Maxima and minima are known as the extrema of a function.

Let θ be the semi-vertical angle of the cone.

It is clear that θ ∈ [0, π/2]

Let r, h, and l be the radius, height, and the slant height of the cone respectively.

The slant height of the cone is given as constant.

 

Now, r = l sinθ and h = l cosθ

The volume (V) of the cone is given by,

V = 1/3πr²h

= 1/3π (l² sin² θ) (l cos θ)

= 1/3π l³ sin² θ.cosθ

Therefore,

dV/dθ = π l³/3 [sin² θ (- sinθ) + cosθ (2sin θ cosθ)]

= π l³/3 [- sin³ θ + 2 sinθcos³θ

d²V/dθ² = π l³/3 [- 3sin² θcosθ + 2cos³θ - 4sin² θcosθ]

= πl³/3 [2cos³θ - 4sin² θcosθ]

Now,

dV/dθ = 0

π l³ / 3 [2 cos³ θ + 2 sin θ cos³ θ] = 0

⇒ sin³ θ = 2sinθ cos ³θ

⇒ tan² θ = 2

⇒ tanθ = √2

θ = tan^{- 1} √2

When, θ = tan^{- 1} √2

Then, tan2 θ = 2 or sin² θ = 2cos²θ

Hence, we have:

d²V/dθ² = πl³/3 [2cos³θ - 14cos³θ]

= - 4π l³ cos ³θ < 0 ∀ θ ∈ [0, π/2]

By the second derivative test, the volume (V) is the maximum when θ = tan^{- 1} √2

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is tan^{- 1} √2

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 25

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^{- 1} √2.

Summary:

Hence we have shown that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^{- 1} √2. Maxima and minima are known as the extrema of a function