You are here

Thomas Simpson and Maxima and Minima - Maximizing a polynomial function

Author(s): 
Michel Helfgott

To determine the different values of x, when that of 3x4 - 28ax3 + 84a2 x2 - 96a3 x + 48b4 becomes a maximum or minimum (Example XXII, page 44).

Let f(x) be the given expression. Then f¢(x) = 12x3 - 84ax2 + 168a2 x - 96a3 . The critical points of f(x) stem from the equation 12x3 - 84ax2 + 168a2 x - 96a3 = 0, which is equivalent to x3 - 7ax2 + 14a2 x - 8a3 = 0. By simple inspection we can realize that the latter equation has solutions x = a,x = 2a,x = 4a. Therefore f¢(x) = (x - a)(x - 2a)(x - 4a). When x < a we have x < 2a and x < 4a, so f¢(x) < 0; when x > a but x < 2a we get x < 4a, thus f¢(x) > 0. By the first-derivative test we can conclude that f(x) adopts a local minimum at x = a. In a similar fashion one can prove that f(x) adopts a local maximum at x = 2a and a local minimum at x = 4a.

As a remark ("scholium") after example XXII (page 45, last paragraph), Simpson discusses the function g(x) = 24a3 x - 30a2 x2 + 16ax3 - 3x4. We have g¢(x) = 24a3 - 60a2 x + 48ax2 - 12x3. Factoring out 12, without much effort we get g¢(x)=- 12(x - a)2 (x - 2a). In a small neighborhood of x = a the derivative is positive, so g(x) does not adopt a local maximum or minimum at x = a. But it does adopt a local maximum at x = 2a because for x < 2a the derivative is positive, while for x > 2a the derivative is negative.

Remark: Both f(x) and g(x) are easy to analyze through the first-derivative test since their derivatives can be factored without much difficulty. The above-mentioned examples may seem rather ad hoc, nonetheless they illustrate quite well how an important test works.

Michel Helfgott, "Thomas Simpson and Maxima and Minima - Maximizing a polynomial function," Loci (August 2010)

Dummy View - NOT TO BE DELETED