# Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base

**Solution:**

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

Maxima and minima are the maximum or the minimum value of a function within the given set of ranges.

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

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

V = 1/3πr^{2}h

⇒ h = 3V/πr^{2}

The surface area (S) of the cone is given by,

S = π r l, where l is the slant height

Hence,

S^{2} = π^{2} r^{2} l^{2}

= π r^{2} ( h^{2} + r^{2})

= π^{2} r^{2} [ (9 V^{2} / π^{2} r^{4}) + r^{2}

S^{2} = (9 V^{2} / r^{2}) + π^{2} r^{4}, differentiating wrt 'r',

2S ds/dr = (-18 V^{2} ) / r^{3} + 4 π^{2} r^{3}

For maximum or minimum,

dS/dr = 0

⇒ (-18 V^{2} ) / r^{3} + 4 π^{2} r^{3} = 0

⇒ 4 π^{2} r^{3} = (18 V^{2} ) / r^{3}

⇒ 2 π^{2} r^{6} = 9V^{2} = 9 x 1/9 (π^{2} r^{4} h^{2}}

⇒ 2 r^{2} = h^{2}

⇒ h = √2 r

⇒ d^{2}S / dr^{2} = 12 π^{2} r^{2} + (54 V^{2} / r^{4})

For all values of V and r.

Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to √2 times the radius of the base

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 24

## Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base

**Summary:**

Here we have shown that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base

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