Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin^{- 1} (1/3)

Solution:

Maxima and minima are known as the extrema of a function

Let r be the radius, l be the slant height and h be the height of the cone of given surface area S.

Also, let a be the semi-vertical angle of the cone.

Then,

S = π r l + π r²

l = (S - πr²) / π r ....(1)

Let V be the volume of the cone.

Then

V = 1/3 π r²h

V² = 1/9 π²r⁴h²

= 1/9π²r⁴ (l ² - r²)

[∵ l ² = r² + h²]

= 1/9π²r⁴ [(S - π r²) / (π r) - r²]

= 1/9π²r⁴ [(S - π²r² - π²r⁴) / (π r) - r²]

= 1/9r² [S² - 2Sπ r²]

= 1/9Sr² (S² - 2π r²) ....(2)

Differentiating (2) with respect to r, we get

2V dV/dr = 1/9Sr² (S² - 2πr²)

For maximum or minimum,

put dV/dr = 0

⇒ 1/9S (2Sr - 8πr³) = 0

⇒ 2Sr - 8πr³ = 0

⇒S = 4πr²

⇒ r² = S/4π

Differentiating again with respect to r, we get

2V d²V/dr² + 2(dV/dr)² = 1/9S (2S - 24πr²)

2V d²V/dr² = 1/9S (2S - 24π x S/4π)

[dV/dr = 0 and r² = S/4π]

2V d²V/dr² = 1/9S (2S - 6S)

2V d²V/dr² = 4/9 S² < 0

Thus, V is maximum when, S = 4πr²

Therefore,

S = π r l + π r²

⇒ 4πr² = π r l + π r²

⇒ 3π r² = π r l

⇒ l = 3r

Now, in ΔCOB,

sin a = OB/BC

= r/l

= r/3r

= 1/3

a = sin^{- 1} (1/3)

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 26

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin^{- 1} (1/3)

Summary:

Hence we have shown that the semi-vertical angle of right circular cone of the given surface area and maximum volume is sin^{- 1} (1/3). Maxima and minima are known as the extrema of a function