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प्रश्न
Show that the right circular cone of least curved surface and given volume has an altitude equal to `sqrt2` time the radius of the base.
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उत्तर
Let the radius of the cone = r
Height of the cone = h
`therefore V = 1/3 pir^2 "h"` = constant quantity
`therefore r^2h = (3 xx "constant quantity")/pi = k` (assumed)
`r^2h = k, h = k/r^2` ...(1)
Curved surface S = `pirl = pir sqrt(h^2 + r^2)`
S = `pir sqrt(k^2/r^4 + r^2)`
`= pir sqrt((k^2 + r^6)/r^4)`
`= pi/r sqrt(k^2 + r^6)`
On differentiating,
`therefore (ds)/(dr) = pi [((6r^5)/(2sqrt(r^6 + k^2)) xx r - sqrt(r^6 + k^2) * 1)/r^2]`
`= pi * (3r^6 - (r^6 + k^2))/(r^2 sqrt(r^6 + k^2))`
`= (2r^6 - k^2)/(r^2 sqrt(r^6 + k^2))`
For maximum and minimum, `(dS)/(dr) = 0`
`=> 2r^6 - k^2 = 0`
`=> r^6 = k^2/2`
`=> r^6 = (h^2 r^4)/2` ...(2)
⇒ h2 = 2r2
∴ h = `sqrt2 r`
At h = `sqrt2 r`, as r passes through `sqrt2 r`
`(dS)/(dr)` changes from -ve to +ve.
∴ S is minimum when h = `sqrt2 r`
Therefore, the height of the right circular cone with the minimum curved surface is `sqrt2` times the radius.
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