### Maximization

In this section Hudde’s general approach and his first two examples will be explained carefully.

To find the \(x\)-value of the maximum for \(f(x) = z,\) Hudde first removed all constant terms, including the maximum value \(z,\) since these do not affect the \(x\)-value at which the maximum occurs. He then transformed the rest of the expression with an arithmetic progression. The root of the transformed equation is the value at which the maximum occurs.

Hudde gave an example

\[3ax^3-bx^3-\frac{2ab^2}{3c}x+a^2b=\,{\rm{some\,\,maximum}}.\]

In this example he used the progression \(0, 1, 2, 3,\dots.\) The result is

\[9ax^3-3bx^3-\frac{2ab^2}{3c}x=0.\]

Dividing by \(x,\) this yields

\[9ax^2-3bx^2-\frac{2ab^2}{3c}=0,\]

which he gave as the answer. He assumed the reader knows how to solve this. He gave neither the value of \(x\) nor of \(z\) (nor did he verify the value of \(z\) is, in fact, a maximum).

There is no evidence Hudde tried any specific values in his examples, but it is likely he usually assumed \(a, b, c\) and \(x\) are all positive. If we further assume to be positive the coefficient, \(3a-b,\) of \(x^3\) in the original equation, thus avoiding complex numbers, then we would have

\[x=\frac{b\sqrt{2a}}{3\sqrt{c(3a-b)}}.\]

Using our modern techniques we can see this corresponds to a local minimum. In fact, the local maximum occurs at the negative of this value. We know Hudde was aware of negative solutions, calling them “false solutions,” as did Descartes.

### Quotient Rule

Most of the rest of Hudde’s examples have \(x\) in the denominator. For a rational function \(\frac{f(x)}{g(x)},\) we would use the quotient rule to find the derivative and set the result equal to \(0.\) Usually this means the numerator is \(0,\) i.e.,

\[f'(x)g(x)-f(x)g'(x)=0.\]

Hudde’s approach with arithmetic progressions leads to an expression more like

\[[x\,f'(x)]g(x)-f(x)[xg'(x)]=0.\]

However, the factor of \(x\) is again easily removed.

### Example with a Denominator

Hudde’s first quotient example is

\[\frac{4a^2b^3+5a^3x+x^5}{x^3}-ax+bx+ab=z.\]

He removed \(ab\) and \(z\) since they do not affect the \(x\)-value for the maximum. He then put the rest over a common denominator:

\[\frac{4a^2b^3+5a^3x+x^5-ax^4+bx^4}{x^3}.\]

Although Hudde didn’t use the notation of negative exponents, we could think of this as

\[4a^2b^3x^{-3}+5a^3x^{-2}+x^2-ax+bx,\]

which would help us understand Hudde’s next step, applying the progression \(-3, -2, -1, 0, 1, 2,\dots\) to the numerator to get

\[-12a^2b^3-10a^3x+2x^5-ax^4+bx^4.\]

Set this equal to \(0\) and solve to find the solution. Hudde left this to the reader!

Hudde’s next examples expanded on this. I refer the interested reader to [Curtin].