Ferrari to Tartaglia, 1548, Question 17

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John Harris

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I've been reading The History of Mathematics: An Open University Course Reader by John Fauvel and Jeremy Gray and wandered out of my depth. I don't know how to approach a problem on page 257, I'm not sure what notation to adopt or how to determine a maximum.

I've jotted down "a+b=8" and "maximize ab(a-b)".

I've expanded "ab(a-b)" to "a(8-a)^2 - a^2(8-a) = c".

I've multiplied it all out and got "2a^3 -24a^2 +64a -c = 0".

What should I have been doing instead?
 
Oh... the problem. Yes.

"Divide eight into two parts such that their product multiplied by their difference comes to as much as possible, proving everything".
 
John Harris said:
I've been reading The History of Mathematics: An Open University Course Reader by John Fauvel and Jeremy Gray and wandered out of my depth. I don't know how to approach a problem on page 257, I'm not sure what notation to adopt or how to determine a maximum.

I've jotted down "a+b=8" and "maximize ab(a-b)".

I've expanded "ab(a-b)" to "a(8-a)^2 - a^2(8-a) = c".

I've multiplied it all out and got "2a^3 -24a^2 +64a -c = 0".

What should I have been doing instead?
Click to expand...

You are more or less correct!

c = -2[a3 -12a2 + 32a]

Have you taken any calculus? For exact answer, you'll have to use calculus.

You can also graphically estimate the maximum.

If the answers are constrained to be integers then you can find that by brute force. According to my graph the maximum is at around 6.4.
 
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Subhotosh Khan said:
You are more or less correct!

c = -2[a3 -12a2 + 32a]

Have you taken any calculus? For exact answer, you'll have to use calculus.

You can also graphically estimate the maximum.
Click to expand...
If I did the calculus correctly (and I am full of turkey and wine so that is questionable), the only exact solution is a conjugate pair in radicals.

Ferrari and Tartaglia knew no calculus so they must have solved it using the cubic formula, which both knew. (In fact, they disputed priority for years about the general solution of the cubic.)

I have no idea how you go about using the cubic formula to find a maximum. Those two were much cleverer than I am. Perhaps they proved somehow that, for a maximum, a and b had to be a particular kind of conjugate pair and then they could use the cubic to find the conjugate solution. Got me stumped.
 
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Thank you for the pointer - I've gone and learned a bit of calculus. I was taught it half a century ago and it has crossed my mind so rarely since that I was uncertain of the rules. I've an answer which makes sense but a follow-up question which stems from it.

maximize ab(a - b) where b = 8 - a

a(8 - a)(a - b) = c
a(8a - 8b -a^2 +ab) = c
a(8a -8(8-a) -a^2 +a(8-a) = c
a(8a - 64 +8a -a^2 +8a -a^2) = c
-2(a^3 - 12a^2 +32a) - c = 0

differentiates to

-2(3a^2 -24a +32) = 0

and the quadratic equation gives two roots,

(24 +- sqrt(192))/6

which is roughly 6.3094 and 1.6913 which add, to check the result, to 8.0007

So the follow-up question - those are rough approximations just to check the answer. What expression would Ferrari have used for his answer? Would it have had sqrt(192) in it? I have my doubts but I can't simplify it (I tried dividing by 36, for example).
 
John Harris said:
Thank you for the pointer - I've gone and learned a bit of calculus. I was taught it half a century ago and it has crossed my mind so rarely since that I was uncertain of the rules. I've an answer which makes sense but a follow-up question which stems from it.

maximize ab(a - b) where b = 8 - a

a(8 - a)(a - b) = c
a(8a - 8b -a^2 +ab) = c
a(8a -8(8-a) -a^2 +a(8-a) = c
a(8a - 64 +8a -a^2 +8a -a^2) = c
-2(a^3 - 12a^2 +32a) - c = 0

differentiates to

-2(3a^2 -24a +32) = 0

and the quadratic equation gives two roots,

(24 +- sqrt(192))/6

which is roughly 6.3094 and 1.6913 which add, to check the result, to 8.0007

So the follow-up question - those are rough approximations just to check the answer. What expression would Ferrari have used for his answer? Would it have had sqrt(192) in it? I have my doubts but I can't simplify it (I tried dividing by 36, for example).
Click to expand...

192 = 64 * 3

So their answer would probablyhave been 4 ± 4√3/3
 
Denis,

Since you and I are both coming from BC era - I am sure you were taught to calculate √2 or √3 by hand!! Those Lamborghini brothers probably used those methods to estimate the values of √3. Or may have left it as √3 as "more exact" answer (I don't think they bothered about cost of significant digits very much - heck they could not even put up a straight tower!!!).
 
Denis said:
SQRT(192) = SQRT(64 * 3) = 8SQRT(3)

What d'heck's Ferrari got to do with it?
Or are you asked: since no calculators then, what guess
would Ferrari have taken?
Since 196 is close nuff to 192, then SQRT(196) = 14;
so you'd get (24 + 14) / 6 = 6.33 (and 1.67): quite close!
Click to expand...
Denis,

I do not think the point is to find an approximation. I think the point is to prove, using the tools available in the 16th century, which means algebra and Euclidean geometry but no calculus, that the exact answer using radicals is, as Subhotosh Khan said,

\(\displaystyle \dfrac{d\{(-2)(a^3 - 12a^2 + 32a - c)\}}{dx} = 0 \implies 3a^2 - 24a + 32 = 0 \implies a = \dfrac{24 \pm \sqrt{24^2 - 4 *3 * 32}}{3 * 2} = 4\left(1 \pm \sqrt{\dfrac{1}{3}}\right).\)

So this method for finding the solution is not following the rules for this puzzle, which really should have been placed under Math Odds and Ends. You LOVE puzzles, denis. Time to get to work. It has me stumped, but it probably depends on the cubic formula (which these two sort of discovered) and on one of the theorems in the later books of Euclid, which few know any longer because we have better tools.
 
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JeffM said:
to prove, using the tools available in the 16th century, which means algebra and Euclidean geometry but no calculus, that the exact answer using radicals is, as Subhotosh Khan said
Click to expand...

The storybooks have the poem which everyone back then agreed would solve the problem, so that's the way I'm going to try. I know it can be done that way and presumably there's nobody here with experience of applying it so I may as well volunteer.

I may be gone for some time.
 
John Harris said:
The storybooks have the poem which everyone back then agreed would solve the problem, so that's the way I'm going to try. I know it can be done that way and presumably there's nobody here with experience of applying it so I may as well volunteer.

I may be gone for some time.
Click to expand...
John

I am going to give you my suggestions in your attempt. A bunch of the poem may relate to the 16th century's formulations of the cubic formula, which, at least in Cardono's version, was stated in multiple forms, always using positive coefficients. This was because he interpreted the problem as involving lengths of the edges of cubes, and lengths are necessarily positive. You can find a bunch of information about the modern version of the cubic formula via Google. This may help parse the poem. And I repeat my guess that part of the 16th century solution related to a theorem from the later books of Euclid, which are not studied today except by historians of mathematics. If Euclid does not work, try Archimedes or Diophantus.

Good luck.

Jeff
 
I wonder whether Douglas Adams ever got to hear of the Depressed Cubic. One can only hope.

The wiki page is fun as far as I've reached but I might need a mug of tea soon.
 
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