# 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 2a4 x = 3(x2 + y2 )2 (2x + 2yy¢). But we have to make y¢ = 0, so 2a4 x = = 3(x2 + y2 )2 (2x). Thus a2/Ö3 = x2 + y2, which in turn becomes a6/3Ö3 = (x2 + y2)3. Since (x2 + y2 ) = a4 x2, we can conclude that a6/3Ö3 = a4 x2, whence

 x = a/ Ö 3Ö3

.

Replacing this value in the original expression we get

 y = a Ö 2/3Ö3 .

An alternative approach, also presented in Doctrine, goes as follows: Taking the cube root we arrive at a4/3 x2/3 = x2 + y2, thus y2 = a4/3 x2/3 - x2. Consequently, 2yy¢ = (2/3)a4/3x-1/3 - 2x, i.e.

y¢ =
 2 3 a4/3 x-1/3 - 2x

2y
.
Making y¢ = 0, it follows that (2/3)a4/3x-1/3- 2x = 0, hence
 x = a/ 4 Ö 27 = a / Ö 3Ö3 .

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

 x = a/ Ö 3Ö3

First of all, let us note that he is undoubtedly working with the "positive part" of y defined by

 y = Ö (a4 x2 )1/3 - x2

x £ 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 = a/ Ö 3Ö3

Hence y attains its greatest value at this point.

Michel Helfgott, "Thomas Simpson and Maxima and Minima - Maximizing y as an implicit function of x," Convergence (August 2010)