+136 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 and their difference is largest. (Hint: Let \(x = a - b\) and express \(abx\) in terms of \(x\) alone.)
\(a =\) ?
\(b=\) ?
+9519 \(a + b = 9\), 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 \(b = \dfrac{9 - 3\sqrt3}2\)
+2401 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/
=^._.^=
