**Of all cones under the same given superficies to find that ( ABD) whose solidity is the greatest** (Example IX, p. 21).

It should be noted that by “superficies” Simpson means “lateral surface”, while “solidity” is an archaic synonym of “volume”. We have \(s = \pi x^2 + \pi xy\), where \(s\) is the lateral surface, \(y\) is the length of the slant side, and \(x\) the radius of the base. Then \(y = \frac{s}{\pi x} - x\). The height \(h\) is found by using the Pythagorean proposition. Thus

\(h = \sqrt{(\frac{s}{\pi x} - x)^2 - x^2} = \sqrt{\frac{s^2}{\pi^2 x^2} - \frac{2s}{\pi}}\),

and consequently

\(V(x) = \frac{1}{3} \pi x^2 \sqrt{\frac{s^2}{\pi^2 x^2} - \frac{2s}{\pi}}\).

Define \(f(x) = \frac{s^2}{9}x^2 - \frac{2 \pi s}{9}x^4\), the square of the volume. Then \(0 = f'(x) = \frac{2s^2}{9}x - \frac{8}{9}\pi s x^3\) implies \(x = \sqrt{\frac{s}{4\pi}}\), which is the value of the radius at which the volume is the greatest. Under these circumstances

\(y = \frac{s - \pi x^2}{\pi x} = \frac{s - \frac{\pi s}{4 \pi}}{\pi \sqrt{\frac{s}{4 \pi}}}= 3\sqrt{\frac{s}{4 \pi}} = 3x \).

**Remark:** As in example IV, Simpson does not mention that the square of the volume is a biquadratic expression. There is a simple algebraic alternative to determine its maximum: define \(z = x^2\), thus the problem is reduced to finding the point where the parabola defined by \(-\frac{2 \pi s}{9} z^2 + \frac{s^2}{9} z\) attains its maximum. Given that this is a quadratic polynomial, we know that this takes place at

\(z = \frac{\frac{-s^2}{9}}{\frac{-4 \pi s}{9}}= \frac{s}{4 \pi}\),

thus \(x = \sqrt{\frac{s}{4 \pi}}\).

*Editor's Note: This article was published in 2005.*