How to find the ratio of height/radius such that surface area of a cone is minimized?

Question

I know that the extrema should be taken at the derivative of $V$, where $${V=\frac{\pi}{3}\cdot r^2\cdot h}$$. But I could only find where it maximizes. That's at h/r is equal to 2.

Answer 1

you can write the formula for $V$ as: $${V = \pi\frac{r^2h}{3}}$$ Differentiate with regards to $r$ to get: $${0 = \frac{\pi}{3} (2rh + r^2\frac{dh}{dr}) \implies \frac{dh}{dr} = \frac{-2h}{r}}$$ Remember this as eqn 1.

Similarly, differentiate the formula of Surface area with regards to $r$ to get: $${\frac{dA}{dr} = \pi(r+\sqrt{h^2+r^2}) + \pi r[1 + (2r + 2h\frac{dh}{dr})\frac{1}{2\sqrt{h^2+r^2}}] = 0}$$ $${\implies r + \sqrt{h^2+r^2} + r[1 + (2r + 2h\frac{dh}{dr})\frac{1}{2\sqrt{h^2+r^2}}] = 0]}$$ $${\implies \sqrt{h^2+r^2} + r\cdot(2r + 2h\frac{dh}{dr})\frac{1}{2\sqrt{h^2+r^2}} = 0}$$ Let this be eqn 2.

Substitute the value of $\frac{dh}{dr}$ from eqn one to get the ratio of $\frac{h}{r}$.

Differentiate eqn 2 again to see if it is a minima or a maxima.

Answer 2

Among all right circular cones of volume $V$, we seek the aspect ratio $\displaystyle{x = \dfrac{h}{r}}$ such that the the total enclosing area, $A$, is minimized.

$V = \dfrac{\pi x r^3 }{3} \longrightarrow r = \left(\dfrac{3 V}{\pi x}\right)^{\dfrac{1}{3}}$

Let $s = \sqrt{r^2+h^2} = r\sqrt{1+x^2}$

Then the area can be expressed in terms of $x$: $$A = \pi r (r + s) = \pi r^2 \big(1 + \sqrt{1+x^2}\big) = \pi \left(\frac{3 V}{\pi x}\right)^{\dfrac{2}{3}} \big(1 + \sqrt{1+x^2}\big)$$

To minimize the area, we need to solve $\dfrac{dA}{dx} = 0$.

$$\therefore \dfrac{dA}{dx} = -\frac{2}{3}\large\big(\small x^{\small\cfrac{-5}{3}}\large\big)\small \cdot\big(1 + \sqrt{1+x^2}\big) + \frac{x^{\cfrac{1}{3}}}{\sqrt{1+x^2}} = 0$$

Simplifying the expression, we obtain:

$$\frac{x^2}{\sqrt{x^2 + 1}} - \dfrac{2}{3}\big(1 + \sqrt{1 + x^2} \big) = 0$$

Credits to Boojum for evaluating the solution: $\displaystyle{x = \dfrac{h}{r} = 2\sqrt{2}}$

Noteworthy points:

The height of the minimal-area cone equals the diagonal of a square enclosing its base.

We could set $V = 1$ without loss of generality.

If the base (included in the area) is a disk of radius $r$, the element (side) length is $s = 3r$.

With apex half-angle $\theta =$ arcsin ${r/s}$, the conical part would be cut from a disk of radius $3r$ with central angle $2\pi$ sin $\theta = 360° * r/s = 120°$.