Inspired by incomplete Muirhead

Algebra Level 5

a 2 b c + b 2 c d \large a^2 bc + b^2 cd

Given that a , b , c a,b,c and d d are positive reals satisfying a + b + c + d = 5 a+b+c+d=5 . If the maximum value of the expression above can be expressed as x y z \dfrac x{y^z} , where x , y x,y and z z are positive integers with y y prime , submit your answer as x + y + z x+y+z .

If you get that no such maximum exists, enter 490023 490023 as your answer.


The answer is 490023.

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1 solution

Manuel Kahayon
May 18, 2016

We attempt use lagrange multipliers to solve this problem. Let f ( a , b , c , d ) = a 2 b c + b 2 c d f(a,b,c,d) = a^2bc+b^2cd and g ( a , b , c , d ) = a + b + c + d 5 g(a,b,c,d) = a+b+c+d-5 . We then need to find a point in 4-dimensional space with positive coordinates with sum 5 in which a stationary point is obtained

f ( a , b , c , d ) ) λ ( g ( a , b , c , d ) ) = a 2 b c + b 2 c d λ ( a + b + c + d 5 ) = 0 f(a,b,c,d))-\lambda(g(a,b,c,d))= a^2bc+b^2cd- \lambda(a+b+c+d-5)=0 . Differentiating both sides with respect to a , b , c , d a,b,c,d respectively gives us λ = 2 a b c = a 2 c + 2 b c d = a 2 b + b 2 d = b 2 c \lambda = 2abc=a^2c+2bcd=a^2b+b^2d=b^2c

Solving for the ratios of a , b , c , d a,b,c,d to each other gives us b = 2 a , d = 3 a 4 , c = 5 a 4 b=2a, d=\frac{3a}{4}, c=\frac{5a}{4} . This gives us ( a , b , c , d ) = ( 1 , 2 , 5 4 , 3 4 ) (a,b,c,d) = (1,2,\frac{5}{4},\frac{3}{4}) . Substituting this into the original equation, we get that the value at the stationary point is 25 2 2 \frac{25}{2^2} .

But, since ( 0.1 , 2.9 , 1 , 1 ) (0.1, 2.9, 1,1) yields a higher value than the value at the turning point, we see that the turning point cannot be a maxima. This implies that there is no maxima for this range of values. Indeed, the maxima can only be obtained on the set of nonnegative real numbers.

So, our answer is 490023 \boxed{490023}

This answer appears to be incorrect. The point you found does not satisfy your equation 2 a b c = a 2 c + 2 b c d 2abc=a^2c+2bcd .

Otto Bretscher - 5 years ago

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Ohh, really? Sorry, let me recheck it... It's actually my first time doing lagrange multipliers so I might have not created a correct solution...

Manuel Kahayon - 5 years ago

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You did the Lagrange multipliers right but then you made an algebra mistake. It's a tricky problem... you will have fun with it. I let you find the answer yourself.

Otto Bretscher - 5 years ago

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@Otto Bretscher Is the point ( 1 , 2 , 5 4 , 3 4 ) (1,2,\frac{5}{4}, \frac{3}{4}) ?

Manuel Kahayon - 5 years ago

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@Manuel Kahayon Yes that's the solution for your equations λ = . . . \lambda=... But here is the next question: Does the function attain its maximum there?

Otto Bretscher - 5 years ago

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@Otto Bretscher Lagrange multipliers are nice! I actually posted this problem to find any algebraic solutions for this... But since I couldn't bring myself to post a problem without writing a solution I looked up lagrange multipliers... I agree, this problem is fun!

Manuel Kahayon - 5 years ago

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@Manuel Kahayon The fun is yet to come. See, we are not done yet. Does the function really attain its maximum there?

Otto Bretscher - 5 years ago

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@Otto Bretscher Hmm... I'm guessing that this can be compared to a 2-dimensional graph of a function f(x) and we are solving for the turning point of the graph, except that we have 4 variables, f(a,b,c,d). Now, we can see that in the original equation, ( 0 , 0 , 0 , 0 ) (0,0,0,0) is also a solution therefore 0 , 0 , 0 , 0 0,0,0,0 is also a "turning point" of sorts, or a local minima\maxima of sorts. Now, by guesswork (or intuition) we can see that if a , b , c , d a,b,c,d are all positive then there must be a local maxima, because if there were a local minima it would imply that the function can reach infinity with only positive a , b , c , d a,b,c,d , which is absurd. So, we also conclude that ( 0 , 0 , 0 , 0 ) (0,0,0,0) is a local minima, since with negative a , b , c , d a,b,c,d we can reach arbitrarily large values of f ( a , b , c , d ) f(a,b,c,d) . So, the function does not attain its maximum there, only its local maximum.

Am I right? I am not that sure if this is a valid solution ....

Manuel Kahayon - 5 years ago

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@Manuel Kahayon But ( 0 , 0 , 0 , 0 ) (0,0,0,0) cannot be a value of ( a , b , c , d ) (a,b,c,d) . So... Please wait until I think up why this thing is not working like it should be TT.TT

Manuel Kahayon - 5 years ago

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@Manuel Kahayon This is a constraint optimization problem, the constraints being a + b + c + d = 5 a+b+c+d=5 and the variables required to be positive. Let me give you a simple analogy: Find the maximal value of x 2 + y 2 x^2+y^2 for positive x x and y y with x + y = 1 x+y=1 .... there is no solution. You would get a maximum, though, if you would allow the variables to be non-negative.

Otto Bretscher - 5 years ago

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@Otto Bretscher Ohh, now I see... The answer is actually obtained when a = 0 a=0 and by using AM-GM on b , c , d b,c,d . Then that means this problem is actually flawed... TT.TT But, please enlighten me, why did we get a turning point on that specific point...?

Then, I'm probably gonna have to delete this problem soon...

Manuel Kahayon - 5 years ago

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@Manuel Kahayon Yes, there is no maximum. You could add a sentence: "Enter 490023 if you come to the conclusion that no maximum is attained"... and the problem will be fine.

In single-variable calculus, when you find a point where the derivative is zero, it may be a maximum, a minimum, or neither (like x 3 x^3 at x = 0 x=0 ... it's the same here

Otto Bretscher - 5 years ago

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@Otto Bretscher Oh, wow, this is actually a really nice way of saving the problem... But, sir, can you please post a solution for me...? I do not want to proceed further knowing that there is still much I do not know about Lagrange Multipliers...

Manuel Kahayon - 5 years ago

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@Manuel Kahayon You can just modify your solution, you are almost there. Just show that the point you found does not give you a maximum by giving another point where the value is higher.... one more line.

Otto Bretscher - 5 years ago

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@Otto Bretscher Oh, okay, I understand. Thanks for all your help, sir Otto. Thanks to you, I was able to learn how to use Lagrange Multipliers correctly and what these imply... And also, thank you for teaching me how to study a specific topic thoroughly before posting a problem about it... TT.TT This conversation will certainly help me in many ways in the future!

Manuel Kahayon - 5 years ago

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@Manuel Kahayon It was fun chatting with you... you pick things up very quickly. I wish all my students were as inquisitive and perceptive as you are. I wish you a bright future and lots of fun with maths!

PS. No need to call me "Sir"... it sounds so aristocratic and British. You may call me Comrade Otto ;)

Don't forget to modify your solution so that it will show the right answer!

Otto Bretscher - 5 years ago

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@Otto Bretscher Very well, Comrade Otto. I hope that we chat again in the near future...

(This feels so wrong! Calling someone 44 years older than me as a comrade... Nevertheless, I will probably continue calling you Comrade Otto... Sorta feels like something straight out of the USSR :))

Manuel Kahayon - 5 years ago

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@Manuel Kahayon "Comrade" knows no age. I hang out in Cuba a lot, and we are all "compañeros/as", from the kids to the members of the Central Committee.They actually have a Compañero Otto in the Central Committee whom I know ;)

Otto Bretscher - 5 years ago

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@Otto Bretscher I just happened to come across your name again Comrade Otto! I'm not sure where else I would be able to contact you (email perhaps?), so I'm sending a quick message here. My friend bought me a copy of your book on linear algebra. It's the fourth edition. He didn't buy it new, but if I had the money, I definitely would've. Anyway, shortly after I joined this website, I came across your name somewhere on this site, and decided to search you up on the internet. That's where I found your book. Then--I think it was only 2-3 days later--I was watching a lecture on abstract algebra by Benedict Gross (former dean of Harvard, Gross–Zagier Theorem) from 2003, and he mentioned your book!!! So I decided I had to get a copy because I am fascinated by the determinant. I just read section 6.1, and I am very impressed! I had never seen an explanation of determinants like that before. Using the linearity property of determinants to break them up into sparse "pattern matrices" was ingenious! You didn't come up with that, did you? By the way, I found a typo on page 252. The arrow over the y vector is missing.

James Wilson - 3 years, 7 months ago

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@James Wilson Ok, I'm guessing you didn't come up with it, but I still really enjoyed it. I'm looking forward to reading the rest of the chapter.

James Wilson - 3 years, 7 months ago

@Manuel Kahayon Your solution looks great now! Just one small remark about terminology: I would call that point you find a "stationary point" rather than turning point.

Otto Bretscher - 5 years ago

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@Otto Bretscher Oh, thanks, comrade! Better make my solution as close to perfect as possible...

Manuel Kahayon - 5 years ago

@Manuel Kahayon Ahh, now I know! By the same train of thought above we conclude that the point we solved above is the local maxima, but we actually have multiple local minima on all points along the line a = b = 0 , c , d a=b=0, c,d such that c + d = 5 c+d=5 since there is not only a single solution for the system of equations for λ \lambda . All the solutions are either local minima\maxima and it is up to us to determine what the points we obtain are...

But to have local minima along a line? 4-dimensional graphs? My mind is blown...

Manuel Kahayon - 5 years ago

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@Manuel Kahayon We are working on a hyper-plane in 4-space. Once you learn linear algebra, you will see that this is not such a big deal. The real problem is that the domain is not compact ... it would be compact if we would allow non-negative variables. Thus we cannot be sure that the given function will attain a maximum and a minimum.

Otto Bretscher - 5 years ago

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@Otto Bretscher Ohh, now this is enlightening. I'm learning so much! Sir Otto, how do you determine whether a given point is a maxima or a minima?

Manuel Kahayon - 5 years ago

Now I found the flaw. It's in the calculations... Please wait a while...

Manuel Kahayon - 5 years ago

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