To find the greatest value of y in the equation a^{4} x^{2} = (x^{2} + y^{2} )^{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 a^{2}/Ö3 = x^{2} + y^{2}, which in turn becomes a^{6}/3Ö3 = (x^{2} + y^{2})^{3}. Since (x^{2} + y^{2} ) = a^{4} x^{2}, we can conclude that a^{6}/3Ö3 = a^{4} x^{2}, whence
.
Replacing this value in the original expression we get
An alternative approach, also presented in Doctrine, goes as follows: Taking the cube root we arrive at a^{4/3} x^{2/3} = x^{2} + y^{2}, thus y^{2} = a^{4/3} x^{2/3}  x^{2}. Consequently, 2yy¢ = (2/3)a^{4/3}x^{1/3}  2x, i.e.
y¢ = 
2
3

a^{4/3} x^{1/3}  2x 
2y

. 

Making
y¢ = 0, it follows that (2/3)
a^{4/3}x^{1/3} 2
x = 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
First of all, let us note that he is undoubtedly working with the "positive part" of y defined by
y = 
Ö

(a^{4} x^{2} )^{1/3}  x^{2}


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
Hence y attains its greatest value at this point.