Pierre de Fermat (1601 - 1665) and Max-Min Problems

The following basic result in first semester calculus is nowdays often referred to in these texts as Fermat's Theorem:

Fermat's Theorem: If f has a local maximum or minimum at c, and if f '(c) exists, then f '(c) = 0.

Indeed, Fermat's own contemporaries so identified him as having a strong interest in maximum-minimum problems that ina priority dispute over the introduction of coordinate systems and analytic geometry, Rene Descartes sarcastically referred to Fermat as vostre Conseiller De Maximus et Minimis.

However, the rigorous definition of f '(c) using a precise formulation of the concept of limits as we present it nowdays in calculus courses awaited Augustin-Louis Cauchy (1789 - 1857) and Karl Weierstrass (1815 - 1897), with the latter formulating the limit definition in delta-epsilon language just as we teach today. In the 1600's, various scientists including Barrow, Descartes and Fermat solved specific calculus problems somewhat on a case by case basis, but the systematic formulation and broader use of calculus as a general tool awaited the researches of Isaac Newton (1642 - 1727) and Gottfried Wilhelm Leibniz (1646 - 1716).

So let us look at an example of how Fermat himself solved a typical max-min problem, the problem of dividing a straight line segment of length L into two pieces so that the rectangle formed with the two pieces as sides has maximum area. To use more contemporary notation than Fermat did, let x and L - x denote the length of the two sides, x in [0,L], and let f(x) denote the corresponding area of the rectangle. Thus

(1) f(x) = x(L - x) = Lx - x².

Rather than taking the first derivative and solving f '(x) = 0 as we do in basic calculus, Fermat employed the intuitive principle that near a maximum or minimum, the function f(x) changes less rapidly with an increment of x to x + e. Hence, as a first step toward solving this problem, Fermat considered the resulting equation

(2) f(x + e) = f(x).

Calculating explicitly using formula (1), we have

(3) L(x + e ) - (x + e)² = Lx - x²

whence

(4) L - 2x - e = 0.

Now Fermat directed that the e term be discarded (which we might consider as putting e = 0) and so he obtains the solution L = 2x or x = L/2.

If we check with the usual calculus methods, and calculate f '(x) from (1), we have f '(x) = L - 2x, in agreement with (4) in the case that e = 0.