A perfect welcome to 2016
f ( x , y , z , v , w ) = x z + 2 y z + 3 z v + 7 z w
Let M be the maximal value of f if x 2 + y 2 + z 2 + v 2 + w 2 = 2 0 1 6 and all variables are positive. If M = f ( x 0 , y 0 , z 0 , v 0 , w 0 ) , find w 0 .
The answer is 28.
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2 solutions
Moderator note:
Because there is a common term z , this is much easier to approach.
How can we deal with the case of x y + 2 y z + 3 z v + 7 v w ?
Is this solvable by quadratic forms? I got the matrix
⎣ ⎢ ⎢ ⎢ ⎢ ⎡ 0 0 1 / 2 0 0 0 0 1 0 0 1 / 2 1 0 3 / 2 7 / 2 0 0 3 / 2 0 0 0 0 7 / 2 0 0 ⎦ ⎥ ⎥ ⎥ ⎥ ⎤
With eigenvalues of 0 , 0 , 0 , − 2 3 7 , 2 3 7 (confirmed by WolframAlpha ). What next?
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Next you find the eigenvectors associated with the largest eigenvalue, and among those you find the one satisfying the constraint.
In this particular case, the "classical inequalities" might be the best approach, though. I'm not dogmatic... ;)
Happy New Year! It's still 2015 over here, but the Champagne is chilled and we are ready to party!
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HAPPY NEW YEAR!!!
So I should solve B a = 0 , where
B = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ − 2 3 7 0 1 / 2 0 0 0 − 2 3 7 1 0 0 1 / 2 1 − 2 3 7 3 / 2 7 / 2 0 0 3 / 2 − 2 3 7 0 0 0 7 / 2 0 − 2 3 7 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ , a = ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ a b c d e ⎦ ⎥ ⎥ ⎥ ⎥ ⎤
with x = a , y = b , z = c , v = d , w = e . And I get the eigenvector of
[ 7 1 , 7 2 , 7 3 , 7 3 , 1 ] T
What next?
Do I set x : y : z : v : w = 7 1 : 7 2 : 7 3 : 7 3 , 1 ?
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@Pi Han Goh – You are almost done... now you want a scalar multiple of the vector that satisfies our 2016 constraint.
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@Otto Bretscher – I've edited my comment, can you read again? Thanks.
I think I'm right with the ratios, just need your confirmation.
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@Pi Han Goh – Yes, the proportion is equivalent to the scalar multiple approach. Nicely done!
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@Otto Bretscher – Thanks. It's ridiculously tedious (for me) to solve the Cayley Hamilton, det ( A − λ I 5 ) = 0 . How do you solve it manually (without the using of computer assistance)? Same goes with solving for B a = 0 .
Is there a shortcut?
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@Pi Han Goh – In principle, the eigenvalues are hard (perhaps even impossible) to find for matrices of size 5 x 5 and larger. In our example, the task is easy since 0 is a triple root so we are down to a quadratic polynomial.
In principle, the eigenvectors are the easy part... just solving a linear system, by row reduction. The numbers can be messy, though.
Do you approve of my solution? I stepped out of my comfort zone and used "AM-GM" for the first (and perhaps only) time in my life ;) (But I did check my work with the matrices, just the way you did it, to make it "official"... I don't fully trust this "classical inequalities" stuff...it feels so arbitrary)
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@Otto Bretscher – Thank you for your response. Comrade Otto!
Of course I approve it! Haha! I thought this is another quadratic form question, so I treat it as such.
Nicely done.....
I solved it via Lagrange multiplicator algorithm. Already during the substitution of λ together with the restriction equation I got expressions such y(x), z(x), v(x), and w(x). So the original equation reduced by one variable x after derivation quickly delivered the result x=4 from which w=7x=28 followed .
Mr. Bretscher: I wish you a happy new year filled by maximum health paired with a lot of math successes and a harmonic familiar atmosphere. I appreciate your mathematical know-how in special manner. Salute you!!
First we will use Cauchy-Schwarz and AM-GM to find an upper bound: f ( x , y , z , v , w ) = z ( x + 2 y + 3 v + 7 w ) ≤ z 1 + 4 + 9 + 4 9 x 2 + y 2 + v 2 + w 2 = 6 3 ( 2 0 1 6 − z 2 ) z 2 ≤ 1 0 0 8 6 3 Equality is attained when 2 0 1 6 − z 2 = z 2 and x : y : v : w = 1 : 2 : 3 : 7 .
We can conclude that z 2 = 1 0 0 8 = x 2 + y 2 + v 2 + w 2 = x 2 + 4 x 2 + 9 x 2 + 4 9 x 2 = 6 3 x 2 so that x 2 = 6 3 1 0 0 8 = 1 6 and x = 4 and w = 7 x = 2 8 ... a perfect number... a perfect year!