# 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

$$x = \frac{a}{\sqrt{27}} = \frac{a}{3\sqrt{3}}$$.

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.