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Extrema of two-variable function in bounded region

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Find absolute maxima and minima of the function in the given region:

    T(x,y) = x2 + xy + y2 - 6x

    Region: Rectangular plate given by: 0 ≤ x ≤ 5, -3 ≤ y ≤ 3

    2. Relevant equations

    First derivative test, fx =0, fy = 0
    Second derivative test, fxxfyy - fxy2 = ?

    3. The attempt at a solution

    I know I need to check the boundaries of the region, which would be four lines connecting the endpoints (0, -3), (0, 3), (5, -3), (5, 3).

    Do I also need to check for extrema the normal way (i.e., not caring about the region) and then see if any of those points lie within the region..? I have a general idea of what I'm supposed to be doing but I'm not 100% sure.
     
  2. jcsd
  3. Oct 3, 2012 #2
    Re: Extrama of two-variable function in bounded region

    On the boundaries you need to find the extrema of a function of a single variable. Inside the rectangle you need to find the extrema of the full expression. So just find the zeroes of the gradient and check which of those are inside the rectangle.
     
  4. Oct 3, 2012 #3

    HallsofIvy

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    It isn't really necessary to use the second derivative test since whether you have a local max or min is not relevant. Evaluate the function at all possible critical points and compare to see which is largest and which smallest.

    You have shown that the partial derivatives are never 0 in the interior of the region so need to check on boundaries. Since this is a square you have four lines as boundary.

    1) x= 0. [itex]f(0, y)= y^2[/itex]. Find critical points of that for y from -3 to 3.
    2) x= 5. [itex]f(5, y)= 25+ 5y+ y^2- 30= y^2+ 5y- 5[/itex].
    3) y= -3. [itex]f(x, -3)= x^2- 3x+ 9- 6x= x^2- 9x+ 9[/itex].
    4) y= 3. [itex]f(x, 3)= x^3+ 3x+ 9- 6x= x^2- 3x+ 9[/itex].

    And, of course, you will need to check boundaries of those: the corners, (0, -3), (0, 3), (5, -3), and (5, 3). Find the values of the functions at all the critical points and see which is largest and which smallest.
     
  5. Oct 3, 2012 #4

    Ray Vickson

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    You also need to worry about the corners (0,-3), etc. A corner is a local minimum if the directional derivative of f is ≥ 0 for all directions pointing into the region (or along the boundary) from that corner, and it is a local max if the directional derivative is ≤ 0. (These tests are, essentially, what the so-called Karush-Kuhn-Tucker conditions are all about.)

    RGV
     
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