# prove : semi verticle angle of right circular cone of given volume and least curved surface area is cot^-1 (root2) .

Let the semi-vertical angle of the cone be $\mathrm{\theta }$.

Since volume is constant, we can write h in terms of V from (1):-

Using (3) in (2):-

$A=\mathrm{\pi r}\sqrt{{\mathrm{r}}^{2}+\frac{9{\mathrm{V}}^{2}}{{\mathrm{\pi }}^{2}{\mathrm{r}}^{4}}}$

$⇒A=\sqrt{{\mathrm{\pi }}^{2}{r}^{4}+\frac{9{V}^{2}}{{r}^{2}}}$

$⇒\frac{dA}{dr}=\frac{4{\mathrm{\pi }}^{2}{\mathrm{r}}^{3}+9{\mathrm{V}}^{2}\left(-\frac{2}{{\mathrm{r}}^{3}}\right)}{2\sqrt{{\mathrm{\pi }}^{2}{\mathrm{r}}^{4}+\frac{9{\mathrm{V}}^{2}}{{\mathrm{r}}^{2}}}}$

Now, A is maximum or minimum when $\frac{dA}{dr}=0$

2r3+9V2-2r3=0

2r3=18V2r3

2r6=13πr2h2

6=r4h2

2=h2

Hence Proved.

• 24
What are you looking for?