1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Maximizing the function

  1. Nov 19, 2007 #1

    rye

    User Avatar

    1. The problem statement, all variables and given/known data
    max f(x, y) = 2xy + 2y − x^2 − 2y^2
    I need to get the value of x and y and determine if it gives the global maximum by using the infinity approach only.
    Boundary is Negative infinity and Positive infinity


    2. Relevant equations
    max f(x, y) = 2xy + 2y − x^2 − 2y^2


    3. The attempt at a solution
    I am aware of the first step:
    Partial differentiate with respect to x ++++++ 2y − 2x = 0
    Partial differentiate with respect to y ++++++ 2x+ 2− 4y = 0
    Solving both the equations will give x = y = 1

    From here, I do not know how to use f(±INFINITY,±INFINITY) to determine if x = y = 1 yields the global maximum.


    Your help is greatly appreciated. I have an exam next week on this.

    Thanks Rye.
     
  2. jcsd
  3. Nov 19, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If you are using partial derivatives, this is hardly "pre-calculus"! I am going to move to this to calculus and beyond.

    I don't know what your teacher means by "the infinity approach" but you DON'T "use f(±INFINITY,±INFINITY)" such a thing does not exist. You can, however, look at LIMITS as x and y go to infinity. In particular, along the line y= x, the function becomes f(x, x) = 2x2 + 2x − x2 − 2x2= 2x- 2x2. What happens to that as x goes to infinity? I particularly suggested y= x because f(x,y) = 2y- (x-y)2.

    Again, this may not be the "infinity approach" but letting u= x-y, v= x+ y and writing f as a function f u and v might be enlightening.
     
    Last edited: Nov 19, 2007
  4. Nov 19, 2007 #3

    rye

    User Avatar

    Hi,

    Thanks for the reply. What I meant was the functions end points are positive and negative infinity. In order to test if f(1,1) qualifies to be the max value, I have to test out f(+∞,+∞) and f(-∞,-∞).

    I guess what your mean by as x and y approaches infinity makes sense with f(+∞,+∞) and f(-∞,-∞).

    However, I do not know what is the step from here. Pls enlighten. Thanks!
     
  5. Nov 19, 2007 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    NO, what I meant was that f(+[itex]\infty[/itex], +[itex]\infty[/itex]) have no meaning. You may be using them as shorthand for limits but you should be clear about that. Actually, since this is a function of two variables, you need more than just the limits as x and y individually go to infinity. You need to consider the possibility of (x,y) going infinitely far from (0,0) along any path. Have you looked at my last suggestion? If you take u= x-y and v= x+ y, then 2y= v- u, 2x= u+v and 2xy= (v^2- 8y^2)/2 so f(x,y)= 2xy- (x-y)^2= v^2/2- u^2/2- u^2= v^2/2- 3u^2/2. What kind of figure is that?
     
  6. Nov 19, 2007 #5

    rye

    User Avatar

    Hi HallsofIvy,

    I am totally confused. I do not know where did u= x-y and v= x+ y come from ?

    I am stuck at the step where I have got x = y = 1. From here how can I determine that indeed f(1,1) gives the global maximum? I would be grateful if you could tell me step by step.. Plsss

    Tks.

    Rye
     
  7. Nov 21, 2007 #6

    rye

    User Avatar

    anyone can help pls??
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Maximizing the function
  1. Maximizing a Function (Replies: 1)

Loading...