from a handpicked tutor in LIVE 1-to-1 classes

# 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 r^{2}

l = (S - πr^{2}) / π r ....(1)

Let V be the volume of the cone.

Then

V = 1/3 π r^{2}h

V^{2} = 1/9 π^{2}r^{4}h^{2}

= 1/9π^{2}r^{4} (l ^{2} - r^{2})

[∵ l ^{2} = r^{2} + h^{2}]

= 1/9π^{2}r^{4} [(S - π r^{2}) / (π r) - r^{2}]

= 1/9π^{2}r^{4} [(S - π^{2}r^{2} - π^{2}r^{4}) / (π r) - r^{2}]

= 1/9r^{2} [S^{2} - 2Sπ r^{2}]

= 1/9Sr^{2} (S^{2} - 2π r^{2}) ....(2)

Differentiating (2) with respect to r, we get

2V dV/dr = 1/9Sr^{2} (S^{2} - 2πr^{2})

For maximum or minimum,

put dV/dr = 0

⇒ 1/9S (2Sr - 8πr^{3}) = 0

⇒ 2Sr - 8πr^{3} = 0

⇒S = 4πr^{2}

⇒ r^{2} = S/4π

Differentiating again with respect to r, we get

2V d^{2}V/dr^{2} + 2(dV/dr)^{2} = 1/9S (2S - 24πr^{2})

2V d^{2}V/dr^{2} = 1/9S (2S - 24π x S/4π)

[dV/dr = 0 and r^{2} = S/4π]

2V d^{2}V/dr^{2} = 1/9S (2S - 6S)

2V d^{2}V/dr^{2} = 4/9 S^{2} < 0

Thus, V is maximum when, S = 4πr^{2}

Therefore,

S = π r l + π r^{2}

⇒ 4πr^{2} = π r l + π r^{2}

⇒ 3π r^{2} = π r l

⇒ l = 3r

Now, in ΔCOB,

sin a = OB/BC

= r/l

= r/3r

= 1/3

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

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

visual curriculum