Maximizing y as an implicit function of x

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 a^4/3 x^2/3 = x² + y², thus y² = a^4/3 x^2/3 - x². Consequently, 2yy¢ = (2/3)a^4/3x^-1/3 - 2x, i.e.

y¢ = 2 3 a4/3 x-1/3 - 2x 2y .

Making y¢ = 0, it follows that (2/3)a^4/3x^-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.