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

Author(s):
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.