Question

Show that the right circular cone of least curved surface and given volume has an

altitude equal to underroot 2 times the radius of the base.

Answer 1

Right circular cone.
Radius R. of the base.
Altitude =h. Slanting edge = s.

$curved \: surface \: area \: = a = \: \pi \: r \: s \\ a = \pi \: r \: \sqrt{ {r}^{2} + {h}^{2} } \\ volume \: = v = \pi \: {r}^{2} h \: \div 3 = constant \\ \frac{dv}{dr} = 0 \\ = > 2r \: h \: + {r}^{2} \frac{dh}{dr} = 0. \\ \frac{dh}{dr} = - \frac{2h}{r} \\ now \: \frac{da}{dr} = \sqrt{ {r}^{2} + {h}^{2} } + \frac{r \: \times ( \: 2r + 2h \: \frac{dh}{dr} )}{ 2\sqrt{ {r}^{2} + {h}^{2} } } = 0\\ ( {r}^{2} + {h}^{2} ) +r (r - \frac{2 {h}^{2} }{r} ) = 0 \\ 2 {r}^{2} = {h}^{2} \\ h = r \: \sqrt{2} .$
see the derivation above. for the least curved surface area , da/ dr =0.
so we get answer. h = sqrt(2) × r.