Inspired by Comrade Otto Bretscher
f a ( x , y , z ) = a x 2 + 2 a y 2 + 1 5 z 2 + a x y + 2 a y z + 1 5 z x + x + 2 y + 3 z + 1 0
If the function above attains a global minima for some a , submit your answer as 1 0 a .
If the value of a is bounded in the form a ∈ ( p , q ) when the function attains its global minima, submit p + q .
If you come to the conclusion that no global minima exists, enter 666.
The answer is 33.75.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Exactly comrade! I (+1) This is indeed a inspirational solution for how calculus problems are to be dealt by means of algebra. The concept of Hessian as twice the matrix of the quadratic form is great! You mentioned it to me before so I just made the platform to utilize it ;) .
Log in to reply
This is a very well-designed and thoughtful problem indeed. You could have made it even trickier by giving f a a minimum at one endpoint, a = 3 0 , say, but not at the other.
Log in to reply
Yes that would have been more interesting rather. I would try to implement all these in my next problem .Actually here our school syllabus doesn't appreciate creative ideas for maths. They even look for same solution for a same problem , or they won't give marks sufficiently :( it's all messed up here.
Log in to reply
@Aditya Narayan Sharma – Yes, of course, Comrade, this is a huge problem around the world. I think there is a political dimension to this: The capitalist ruling classes realize that they need some people who know some math, as engineers and to come up with ever crazier financial instruments ("derivatives"). But they don't want us to really think : thoughtless people make "better" soldiers, consumers, voters, etc. , more docile and easier to manipulate and distract. Real teaching and real learning is a subversive, revolutionary art, and I see this as my real mission as an educator.
Just consider the current election process in the US: would a thinking person really want to vote for any of these characters, besides Bernie (whom they are now squeezing out)?
One of my heroes and role models is the great Bertrand Russell, who, besides being an eminent philosopher, writer, mathematician, logician, and educator was also relentlessly challenging the system, with the Vietnam Tribunals and otherwise.
How did you immediately know that we need to use f 3 0 ( 0 , − t , 2 t ) and f 1 5 / 4 ( − 2 t , 0 , t ) ? Don't you need to do the kernel thingy? What shortcut did you use this time?
Log in to reply
Yes, Comrade, that's what I'm doing here; those are the kernels!
Yup, I also used the gradient-Hessian method myself, Comrade Otto! This really ought to be classified as a Calculus prob.
Very nice problem, Comrade! Thank you!
Twice the matrix of the quadratic form (the first six terms) is A = ⎣ ⎡ 2 a a 1 5 a 4 a 2 a 1 5 2 a 3 0 ⎦ ⎤
This matrix is positive definite if (and only if) all the leading principal minors are positive: 2 a > 0 , 7 a 2 > 0 and det A = 2 a ( 4 a − 1 5 ) ( 3 0 − a ) > 0 ; this is the case for 4 1 5 < a < 3 0 . The matrix is positive semi-definite for 4 1 5 ≤ a ≤ 3 0 , but we can check that f a ( x , y , z ) fails to have a global minimum when a = 4 1 5 and a = 3 0 due to the linear terms: f 3 0 ( 0 , − t , 2 t ) = 4 t + 1 0 and f 1 5 / 4 ( − 2 t , 0 , t ) = t + 1 0 .
Thus the answer is p + q = 4 1 5 + 3 0 = 3 3 . 7 5
(This solution can be framed in terms of calculus, interpreting A as the Hessian matrix, but the problem is labelled "Algebra")