**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* = p*x*^{2} + p*xy*, where *s* is the lateral surface, *y* is the length of the slant side, and *x* the radius of the base. Then *y* = *s*/p*x* - *x*. The height *h* is found by using the Pythagorean proposition. Thus

\(h = \sqrt{\frac{1}{ (\frac{s}{px} - x)^2 - x^2}} = \sqrt{\frac{1}{\frac{s^2}{p^2 x^2} - \frac{2s}{p}}} \),

and consequently

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

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

\(y = \frac{s - px^2}{px} = \frac{s - \frac{ps}{4p}}{p\sqrt{\frac{1}{\frac{s}{4p}}}} = 3\sqrt{\frac{1}{\frac{s}{4p}}} = 3x\).

**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 -(2p*s*/9)*z*^{2} + (*s*^{2}/9)*z* attains its maximum. Given that this is a quadratic polynomial, we know that this takes place at

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

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