Let $r$ be the radius of the base, $h$ be the altitude, and $l$ be the slant height of the right circular cone. Let $V$ be the given volume and $S$ be the curved surface area.

Step 1: Express volume and curved surface area

Volume of cone:

$$V = \frac{1}{3}\pi r^2 h$$

Since $V$ is given (constant), we have:

$$h = \frac{3V}{\pi r^2}$$

Curved surface area:

$$S = \pi r l, \quad \text{where } l = \sqrt{r^2 + h^2}$$

Step 2: Express $S$ in terms of $r$

Substituting $h$ in $l$:

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

Therefore,

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

Step 3: Minimize $S^2$ (for easier differentiation)

$$S^2 = \pi^2 r^2 \left(r^2 + \frac{9V^2}{\pi^2 r^4}\right) = \pi^2 r^4 + \frac{9V^2}{r^2}$$

Differentiating with respect to $r$:

$$\frac{d(S^2)}{dr} = 4\pi^2 r^3 - \frac{18V^2}{r^3}$$

Step 4: Find critical point

Setting $\frac{d(S^2)}{dr} = 0$:

$$4\pi^2 r^3 = \frac{18V^2}{r^3}$$ $$r^6 = \frac{9V^2}{2\pi^2}$$

Step 5: Find $h/r$ ratio

From $h = \frac{3V}{\pi r^2}$ and $r^3 = \frac{3V}{\sqrt{2}\,\pi}$:

$$\frac{h}{r} = \frac{3V}{\pi r^3} = \frac{3V}{\pi \times \frac{3V}{\sqrt{2}\,\pi}} = \sqrt{2}$$

Step 6: Verify minimum

$$\frac{d^2(S^2)}{dr^2} = 12\pi^2 r^2 + \frac{54V^2}{r^4} > 0 \quad \text{for all } r > 0$$

Since the second derivative is positive, $S$ is minimum when $h = \sqrt{2}\, r$.

∴ Hence proved that the right circular cone of least curved surface area for a given volume has altitude equal to $\sqrt{2}$ times the radius of the base.