# 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 = 1/3πr^{2}h

= 1/3π (l^{2} sin^{2} θ) (l cos θ)

= 1/3π l^{3} sin^{2} θ.cosθ

Therefore,

dV/dθ = π l^{3}/3 [sin^{2} θ (- sinθ) + cosθ (2sin θ cosθ)]

= π l^{3}/3 [- sin^{3} θ + 2 sinθcos^{3}θ

d^{2}V/dθ^{2} = π l^{3}/3 [- 3sin^{2} θcosθ + 2cos^{3}θ - 4sin^{2} θcosθ]

= πl^{3}/3 [2cos^{3}θ - 4sin^{2} θcosθ]

Now,

dV/dθ = 0

π l^{3 }/ 3 [2 cos^{3 }θ + 2 sin θ cos^{3 }θ] = 0

⇒ sin^{3} θ = 2sinθ cos ^{3}θ

⇒ tan^{2} θ = 2

⇒ tanθ = √2

θ = tan^{- 1} √2

When, θ = tan^{- 1} √2

Then, tan2 θ = 2 or sin^{2} θ = 2cos^{2}θ

Hence, we have:

d^{2}V/dθ^{2} = πl^{3}/3 [2cos^{3}θ - 14cos^{3}θ]

= - 4π l^{3} cos ^{3}θ < 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

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

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