history of mathematics - National STEM Centre

Activity 9.3 Fermat and maxima
9 The beginnings of calculus
You will investigate Fermat's method for maxima and minima in Activity 9.3,
moving on to tangents in Activity 9.4. However, Fermat's argument is difficult to
follow; his reasoning is very unclear. The argument in Activity 9.3 is based on
Fermat's writings, but it is unlikely to be the argument that Fermat actually used.
1 Suppose that you have to divide a straight line of length b into two parts so that
the product z of the two parts is a maximum. How should you do it?
Let x be one of the segments of the line b. Then z = x(b — x).
Fermat decided to look at values of z close to ;c, so he replaced x by x + e, where e is
small, to get a new value for the product. Call this new product z'
a Write down an expression for z'.
Fermat was trying to find the maximum value of the number z. He argued that if the
maximum value happens at x, then at x + e the value must be smaller than at x.
Therefore z — z' is always positive.
b Find an expression for z - z' •
Fermat then goes on to use an argument similar to the following. If z - z' > 0 for all
values of e, then e 2 + e(2x -b)>0 for all values of e. This can only occur if the
coefficient of e is zero. Therefore x = -^ b.
c Explain why the coefficient of e has to be zero.
2 Fermat then gives a second, more complicated, example. In this example he still
knows where the maximum value occurs, perhaps from numerical or graphical
experimentation, but he is trying to prove his result. Again he makes a substitution.
Suppose that you are trying to find the maximum value of the expression
z = a 2x - JC 3 as A: varies, and suppose that it happens when x = X. Then
z = a 2X - X 3 . For a value x = X + e, where e is small, the new value of a 2x - x 3
must be smaller than the old one. Call this new value z'.
a Write down an expression for z'.
b Find and simplify an expression for z - z' -
The argument is now more complicated than for question 1. As z - z' > 0 for all
values of e which are small, the value of e(3X 2 — a 2 ) + 3Xe 2 + e 3 must be positive
when e is small. When e is small, e is very small compared with e 2 and can be
ignored. Thus z-z' ~ e[3X 2 — a 2 j + 3Xe 2 . You can now use the argument of
question 1.
c What does the argument of question 1 give for the value of X when z is a
maximum? How do you resolve the difficulty which follows?
d Check your answer to part d by differentiation.
To see how Fermat constructed tangents, work through Activity 9.4. Fermat's
original argument was difficult to follow and virtually none of his contemporaries,
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