A Lengthy Foe
f ( x , y , z , w ) = x 2 + 9 y 2 + 4 z 2 + w 2 − 6 x y + 4 x z − 2 x w − 1 2 y z + 6 y w − 4 z w + x − 3 y + 2 z + w
Find the (global) minimum of f ( x , y , z , w ) .
Enter 666 if you come to the conclusion that no such minimum exists.
Posted at the request of a Comrade.
The answer is 666.
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.
2 solutions
Yes, completing the squares is one good option, Comrade (+1).... let's see what other solutions we get. I labeled this one "Calculus".
Log in to reply
Using Langrange multiplier is one other way to solve this problem.
Log in to reply
Good point! Happy Victory Day (Ngày Chiến thắng), Comrade!
Log in to reply
@Otto Bretscher – Thank sir!
Log in to reply
@Khang Nguyen Thanh – I fondly remember April 30, 1975, one of the happiest days of my life, when Hồ Chí Minh City was liberated .... Vietnam was free and united.
Log in to reply
@Otto Bretscher – Yes, that day is an unforgettable day in Vietnam history. Thank sir one more time for your sentiments towards my country.
Log in to reply
@Khang Nguyen Thanh – My advisor's advisor, the great Alexander Grothendieck, was teaching math in Hanoi (or in the woods nearby, to protect his students from US bombs) during your war of liberation. He thought of this as one of the most important and meaningful things he did in his life.
Log in to reply
@Otto Bretscher – The only thing we need to do for this question is to prove that the determinnant for this matrix has at least one root of 0 right?
⎣ ⎢ ⎢ ⎡ 1 − 3 2 − 1 − 3 9 − 6 3 2 − 6 4 − 2 − 1 3 − 2 1 ⎦ ⎥ ⎥ ⎤
Working : (R2 -> R2 - 3R3) ==> one entire row is all zero ==> determiniant is zero. QED
Log in to reply
@Pi Han Goh – This problem isn't challenging enough for you, Comrade... try this one
I'm a little busy (Sunday morning activities)... I'll be back
Log in to reply
@Otto Bretscher – You didn't answer my question: is my working correct?
Log in to reply
@Pi Han Goh – Your solution is not complete. You could still have a minimum; it depends on the linear terms. If there were no linear terms, for example, the minimum would be 0.
The express ion can be written as : ( x − 3 y + 2 z − w + 2 1 ) 2 + 2 w − 4 1 We see for all ( x , y , z , w ) , f ( x , y , z , w ) ≥ 2 w − 4 1 . so 2 w − 4 1 does not have any minimum which is clear . So no minimum exists.
Alter: We have by first partial differentiation ,
⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 2 x − 6 y + 4 z − 2 w = − 1 1 8 y − 6 x − 6 z + 6 w = 3 8 z + 4 x − 4 w − 6 y = − 2 2 w − 2 x + 6 y − 4 z + 1 = 0
This can be represented in matrix form like this : . ⎝ ⎜ ⎜ ⎛ 2 1 8 8 2 − 6 − 6 4 − 2 4 − 6 − 4 6 − 2 6 6 − 4 ∣ ∣ ∣ ∣ − 1 3 − 2 − 1 ⎠ ⎟ ⎟ ⎞
R 4 → R 4 + R 1 , ⎝ ⎜ ⎜ ⎛ 2 1 8 8 0 − 6 − 6 4 0 4 − 6 − 4 0 − 2 6 6 0 ∣ − 1 ∣ ∣ ∣ 3 − 2 − 2 ⎠ ⎟ ⎟ ⎞
The last row clearly suggests we have an inconsistent set of equations and thus there is no critical point for f ( x , y , z , w ) . so the global minima doesn't exist.
You are on the right track (+1) but the matrix is not quite right: the variables are not aligned properly and some entries are wrong. The coefficient matrix is actually of rank 1.
Log in to reply
Oops. I will correct it as soon as I reach home, actually my phone isn't comfortable for typing so I made the mistake
Log in to reply
No problem. You don't even really need the matrices: We can see that the first and the last of your four equations are inconsistent, as you say.
Log in to reply
@Otto Bretscher – Oh yes we can simply deduce with eqn 4 &1 that there are no critical points. Okay I will fix that soon . anyways I luv this and the previous problem.
Lol.....dangerous matrix
Log in to reply
Typed all these stuff through my speedy mobile :p . I hope u can imagine
Log in to reply
I hope I can help u type that ..in the train ...
Solved it with the same gradient matrix method above. Nice problem, Sir Otto Bee!
Relevant wiki: Completing The Square
We have:
f ( x , y , z , w ) = x 2 + 9 y 2 + 4 z 2 + w 2 − 6 x y + 4 x z − 2 x w − 1 2 y z + 6 y w − 4 z w + x − 3 y + 2 z + w = ( x − 3 y + 2 z − w + 2 1 ) 2 + 2 w − 4 1
If we choose x − 3 y + 2 z − w + 2 1 = 0 , we get f ( x , y , z , w ) = 2 w − 4 1 → − ∞ when w → − ∞
So the global minimum of f ( x , y , z , w ) doesnot exist.