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 = 1/3πr2h
⇒ h = 3V/πr2
The surface area (S) of the cone is given by,
S = π r l, where l is the slant height
Hence,
S2 = π2 r2 l2
= π r2 ( h2 + r2)
= π2 r2 [ (9 V2 / π2 r4) + r2
S2 = (9 V2 / r2) + π2 r4, differentiating wrt 'r',
2S ds/dr = (-18 V2 ) / r3 + 4 π2 r3
For maximum or minimum,
dS/dr = 0
⇒ (-18 V2 ) / r3 + 4 π2 r3 = 0
⇒ 4 π2 r3 = (18 V2 ) / r3
⇒ 2 π2 r6 = 9V2 = 9 x 1/9 (π2 r4 h2}
⇒ 2 r2 = h2
⇒ h = √2 r
⇒ d2S / dr2 = 12 π2 r2 + (54 V2 / r4)
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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