Thomas Simpson and Maxima and Minima - Maximizing y as an implicit function of x

Michel Helfgott

To find the greatest value of y in the equation a4 x2 = (x2 + y2 )3 (Example XX, p. 42).


Assuming that \(y\) is a function of \(x\), implicit differentiation (Simpson says: “by putting the whole equation into fluxions”) leads to \(2a^4 x = 3(x^2 + y^2)^2 (2x + 2yy')\). But we have to make \(y' = 0\), so \(2a^4 x = 3(x^2 + y^2)^2 (2x)\). Thus \(\frac{a^2}{3} = x^2 + y^2\), which in turn becomes \(\frac{a^6}{3\sqrt{3}} = (x^2 + y^2)^3\). Since \((x^2 + y^2) = a^4 x^2\), we can conclude that \(\frac{a^6}{3\sqrt{3}} = a^4 x^2\), whence

\(x = \frac{a}{\sqrt{3\sqrt{3}}}\).

Replacing this value in the original expression we get

\(y = a \sqrt{\frac{2}{3\sqrt{3}}}\).

An alternative approach, also presented in Doctrine, goes as follows: Taking the cube root we arrive at \(a^\frac{4}{3} x^\frac{2}{3} = x^2 + y^2\), thus \(y^2 = a^\frac{4}{3} x^\frac{2}{3} - x^2\). Consequently, \(2yy' = \frac{2}{3} a^\frac{4}{3} x^\frac{-1}{3} - 2x\), i.e.

\(y' = \frac{\frac{2}{3} a^\frac{4}{3} x^\frac{-1}{3} - 2x}{2y}\).

Making \(y' = 0\), it follows that \(\frac{2}{3} a^\frac{4}{3} x^\frac{-1}{3} - 2x = 0\), hence

\(x = \frac{a}{\sqrt[4]{27}} = \frac{a}{3\sqrt{3}}\).

Remark: An aspect not considered by Simpson is whether a maximum is actually attained by \(y\) at

\(x = \frac{a}{3\sqrt{3}}\).

First of all, let us note that he is undoubtedly working with the “positive part” of \(y\) defined by

\(y = \sqrt{(a^4x^2)^\frac{1}{3} - x^2}, x \le a\).

This function adopts the value zero at \(x = 0\) and \(x = a\), and is positive on \((0,a)\). There is just one critical point, namely

\(x = \frac{a}{3\sqrt{3}}\).

Hence \(y\) attains its greatest value at this point.


Editor’s Note: This article was published in 2005.