You have 24 metres of fencing and want to enclose a rectangular plot with the largest possible area. Which dimensions should you choose? A long thin rectangle wastes fencing; a square seems better. But how do you prove it?

Fermat invented a method for questions like this around 1636, more than thirty years before Newton and almost fifty before Leibniz published calculus.

Fermat’s problem

Fermat’s first example was exactly this: divide a segment of length $b$ into two parts, $x$ and $b-x$, so that the product $x(b-x)$ is as large as possible. That is our plot: with 24 m of fencing, two adjacent sides add up to $12$, and the area is

$$A(x)=x(12-x)=12x-x^2,\qquad0

Change one side: the other adjusts to keep the perimeter at 24. The area grows until the square and then shrinks.

The idea of adequality

Fermat noticed something simple: near the maximum, the function hardly changes. At the top of a mountain, a step sideways barely changes your height. Far from the top, a step changes it a lot.

So he compared the value at $x$ with the value at a neighbouring point, $x+e$, where $e$ is a small quantity. At the maximum the two should be “almost equal”. Fermat called this near-equality adaequatio, adequality, a word he took from Diophantus.

Let us follow his steps:

  1. Set the neighbours equal. We write $A(x+e)\approx A(x)$: $$12(x+e)-(x+e)^2\approx12x-x^2.$$
  2. Simplify. Expanding and cancelling what appears on both sides: $$12e-2xe-e^2\approx0.$$
  3. Divide by $e$: $$12-2x-e\approx0.$$
  4. Remove $e$. Now that there is no more division, Fermat simply erased the remaining $e$: $$12-2x=0\quad\Longrightarrow\quad x=6.$$

The best plot is the square of side 6, with area 36 m².

Notice that the expression in step 3, once $e$ is erased, is exactly what we now call the derivative: $A'(x)=12-2x$. The maximum is where the derivative vanishes. Fermat was doing differential calculus without the name.

A proof without calculus

Fermat’s method finds the candidate for the maximum. To prove it really is the maximum, there is an algebraic trick: completing the square.

$$A(x)=12x-x^2=36-(x-6)^2.$$

A square is never negative, so $(x-6)^2\ge0$ and $A(x)\le36$. Equality happens only at $x=6$. No other rectangle with a 24 m perimeter has a larger area.

The box of greatest volume

The method also works for harder problems than the plot of land. A factory has square steel sheets, 6 m on a side, and wants to make open boxes: it cuts a square of side $x$ from each corner and folds the flaps up. The base has side $6-2x$, the height is $x$, and the volume is

$$V(x)=(6-2x)^2x=4x^3-24x^2+36x,\qquad0If $x$ is very small, the box is shallow; if it is large, the base shrinks. By adequality, we set $V(x+e)\approx V(x)$, cancel what appears on both sides, divide by $e$ and erase the terms in which $e$ remains. We get

$$12x^2-48x+36=0\quad\Longrightarrow\quad x^2-4x+3=0\quad\Longrightarrow\quad x=1\ \text{or}\ x=3.$$

With $x=3$ there is no base left at all, and the volume is zero. The maximum is at $x=1$: a box with a $4\times4$ base and height 1, with a volume of $16$ m³. For comparison, with $x=0.5$ the volume is $12.5$ m³, and with $x=1.5$ it is $13.5$ m³.

The quarrel with Descartes

Fermat used the same method to find tangents to curves. When the method reached Descartes in 1638, he attacked it: he said it was confused and only worked by luck. Fermat replied calmly, and the two exchanged increasingly sharp letters. In the end Descartes grudgingly admitted the method worked. He simply did not understand why, since Fermat did not explain it.

The most delicate step is step 4: dividing by $e$ assuming $e\neq0$ and then erasing it as if it were zero. It is the same problem Newton would have with his $o$ and Leibniz with his $dx$. Only in the 19th century, with the idea of a limit, did the step become rigorous.

Fermat never published the method. The text, Method for finding maxima and minima, circulated in manuscript copies sent to Mersenne around 1636 and passed from hand to hand among the mathematicians of Paris. It was only printed in 1679, in the Varia opera mathematica edited by his son Samuel, and received a complete edition at the end of the 19th century, in the Œuvres de Fermat by Paul Tannery and Charles Henry.

Exercise: what is the largest area for a perimeter of 40?

The sides are $x$ and $20-x$. The area is $20x-x^2=100-(x-10)^2$, at most 100, for the square of side 10.

Exercise: a plot lies against a wall, so only three sides need fencing. With 24 m of fencing, what is the largest area?

If the two sides perpendicular to the wall measure $x$, the side parallel to it measures $24-2x$. The area is $A=x(24-2x)=24x-2x^2$. By adequality: $24-4x=0$, so $x=6$. The plot is $6\times12$ with area 72 m². This time the best shape is not a square.

Sources: Fermat biography, MacTutor.

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