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Finding the minimum of a 2 variable function.

  1. Dec 14, 2013 #1
    1. The problem statement, all variables and given/known data

    ind the absolute minimum value of the function f(x,y) = 6 + 3xy -2x- 4y on the set D. D is bounded by the parabola y=x^2 and y=4

    2. Relevant equations

    Partial derivatives and a theorem where if F is continuous on a closed bounded set in R(2) then F has an abs. max and abs. minimum

    3. The attempt at a solution
    3
    I found the partial derivative in accordance to x and y:

    Fx=3y-2
    Fy=3x-4

    I found the critical points ( when Fx and Fy = 0) being (4/3 3/2). However, I get stuck when sketching the graph. I am used to sketching a square/ rectangle ( with the boundaries) and dividing it into 4 lines and solving for each with my function. In this I get:

    F(x,x^2)= 6+x^3-2x-4x^2 but then I don't know how to find the smallest value for that, as my x is good for all real numbers no?
     
  2. jcsd
  3. Dec 14, 2013 #2
    You are right that only critical point is x = 2/3 y = 4/3. Surely you can sketch y = ##x^2## and y = 4?

    Is (2/3,4/3) in this area or not? If so, it may represent a min, a max or a saddle. Do you know how to determine which you have?

    If it is a max or a saddle, your min will definitely occur on the boundary, and we can get back to that in the next step.

    Likewise, if it is a min, the absolute min may still occur on the boundary, or maybe this critical point is it. We have to check that.

    So if you can get this first step done, we can look next at the boundary situation.
     
  4. Dec 14, 2013 #3
    hum, 2/3 and 4/2 does lie in those 2 boundaries, I mean the parabola is from (-16,4) to (16,4) ( If I graphed it). However I never took this approach to those problem ( which leads me to ask why we ever found the critical point as we didn't use them in our answer).

    We usually sketched the "region" the describes the boundaries and solve for the function when either the x or y is constant and then use the boundary to find the lowest or highest value.

    As for this example, I think it is just a saddle, but it's a "guestimate". I mean if I plug in my 2 "furthest point" (-16,4) and (16,4) I get a lower / higher point. Don't know if I am looking at this the right way?
     
  5. Dec 14, 2013 #4

    Dick

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    Your work is riddled with errors. The critical point is at (2/3,4/3) not (2/3,4/2) or (2/3,3/2) both of which you've stated. And neither (-16,4) nor (16,4) is even in your region. Draw a graph. What are the real x values that bound your region? As brmath said, if you can show that's a min using the second derivative test then you don't need to check the boundary. It's not (what is it?). So you do need to check the boundaries. That f(x,x^2)= 6+x^3-2x-4x^2 is also wrong for the parabola boundary. Correct that too. What's the function along y=4?
     
  6. Dec 14, 2013 #5
    First, it was a typo, I did get the critical point you just pointed out so my apologies for that.

    If you draw the graph out the parabola does meet the line y=4, I meant the parabola meets the line at x=-2 and x=2 ( my bad also on that, I squared 4 for some reasons...)

    Again, the only method I used in my cal 3 class to find the abs. minimum and max was to draw a "graph" with the boundaries on it.. So it would give me , in all the cases we covered, a rectangle / square where every vertices would be one of the boundaries... So for instance if 0</= x </= 2 and 1</=y</= 2 I would have
    (0,1), (0,2), (2,1),(2,2). After, I would separate the rectangle in 4 lines, and then solve for on every line for which variable is constant....

    SO in THIS case, I did it with the fact that I have 2 lines after sketching my boundaries. L1 : y=4 and L2 y=x^2
    so for L2: f(x,x^2)= 6+3x^2-2x+4x^2... That was how I was taught...

    Then l1: f(x,4)= 6+12x-2x+16... But this is the only method I was taught. I don't even know what is the point of finding my critical point as it looks to be a saddle point. (in this case)
     
  7. Dec 14, 2013 #6

    Dick

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    Indeed it is a saddle point. If you don't know the second derivative test then ok, then just check the value at your critical point f(2/3,4/3) and the minima along your boundary curves. But those have errors in them as well. Are you even sure your original function is f(x,y)=6+3xy-2x-4y? Or is there a typo there as well? It's really hard to work around all this error noise.
     
  8. Dec 14, 2013 #7
    oh yeah the second derivative test, I remember, whoops ( been forever and reviewing for finals). As for the rest of the info, they are right.

    As for the 2nd derivative test, it only applies to local max and min , in this question I'm being ask for absolute min. So I don't think I can use it ( correct me if I'm wrong)
     
  9. Dec 14, 2013 #8

    Dick

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    You don't HAVE to use the 2nd derivative test. You do HAVE to use the critical point somehow. The minimum could be either along the boundary or at the critical point. IF you know the critical point is a saddle point using the 2nd derivative test, then you would know the minimum is not at the critical point. But you can't know that until you find the critical point and show it is a saddle point. Get me? You do have to always find critical points. You can't generally know they aren't what you are looking for until you find them.
     
    Last edited: Dec 15, 2013
  10. Dec 15, 2013 #9
    In this case, when I apply the second derivative test, I get D as a constant ( I don't know if I am doing it wrong..)
    Fx= 3y-2 therefore Fxx= 0 and samething goes for y. So I assumed my critical point was a saddle, afterwards however, I don't know what other points to use to find my abs. max and min.
     
  11. Dec 15, 2013 #10
    Whether there is a max, min or saddle at the critical points, you do need to check the boundary. A min at a critical point can be a local min, so it may be that the abs min is on the boundary.

    To check the boundary, you have 2 cases: y = ##x^2## and y = 4. Each case allows you to express f as a function of 1 variable, so you can find the minimum using ordinary 1 dim calculus. Whichever minimum is smaller is your absolute min.

    Re your second derivative test, you should revisit the criteria. If D > 0 you have either a min or a max depending on whether the Jacobian matrix is positive or negative definite. If D < 0 you have a saddle. If D = 0 the test fails -- you can't conclude anything from it.
     
  12. Dec 15, 2013 #11

    Dick

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    ##f_{xy}## isn't zero. But yes, D is constant and negative. it's a saddle. Now check the boundary! Your min must be there.
     
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