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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," Loci (August 2010)

Dummy View - NOT TO BE DELETED