Problem of Tartaglia (1500-1557): among all positive numbers a, b whose sum is 9, find those for which the product of the two numbers...

$a + b = 9$, so $b = 9 - a$.

Now, $ab(a - b) = a(9 - a)(a - (9- a)) = a(9 - a)(2a - 9) = -2a^3 + 27a^2 - 81a$.

When ab(a - b) is the largest, d(ab(a - b))/da = 0.

$\dfrac{d}{da} (-2a^3 + 27a^2 - 81a) = 0\\ -6a^2 + 54a - 81 = 0\\ 2a^2 - 18a + 27 = 0\\ a = \dfrac{9 \pm 3 \sqrt 3}2$

For brevity, let $f(a) = -2a^3 + 27a^2 - 81a$. 

$f''\left(\dfrac{9 + 3\sqrt 3}2\right) = -12\left(\dfrac{9 + 3\sqrt 3}2\right) + 54 = -18\sqrt3 < 0$

$f''\left(\dfrac{9 - 3\sqrt 3}2\right) = -12\left(\dfrac{9 - 3\sqrt 3}2\right) + 54 = 18\sqrt3 > 0$

So, the maximum value of ab(a - b) is attained at $a = \dfrac{9 + 3\sqrt3}2$.

Since b = 9 - a, $b = \dfrac{9 - 3\sqrt3}2$ when ab(a - b) is the largest.

Therefore, $a = \dfrac{9 + 3\sqrt3}2$, and $b = \dfrac{9 - 3\sqrt3}2$

Your answer looks very cool, but I wanted to try to see if it can be solved with calculus too because I haven't learned it yet. :))

m = maximum of ab(a-b)

x = a - b (like the hint said)

9 = a + b

a = (9+x)/2

b = (9-x)/2

m = maximum of (9+x)/2*(9-x)/2*x

Note: x has to be be nonegative since if b > a, and a and b are positive integers, ab(a-b) will be negative.

This is a cubic with roots at -9, 0, and 9.

By plugging in points and drawing a sketch, we can get an idea of what the cubic looks like. 

The graph approaches 0 from the positive side from as x approaches -9 from the negative side. 

Then, the graph crosses the x axis at -9, and goes down, but then goes back up to the x axis at 0. 

After that, the graph dips up but returns back to 0 at x = 9, and just continues going down.

So our maximum is at the turning point between 0 and 9, this is the distance m. 

Then we can use the method written here:

https://www.themathdoctors.org/max-and-min-of-a-cubic-without-calculus/

=^._.^=