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Absolute extrema

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Find absolute extrema of the function over the region R. f(x,y) = 3x2 +2y2 -4y, R: the region in the xy plane bounded by the graphs of y=x^2 and y=4


    2. Relevant equations
    second partial derivative test
    d=fxxfyy-fxy2


    3. The attempt at a solution
    This is my practice test for the final next week and I'm really unsure how to do problems like this one. I can find the critical points easily enough
    fx= 6x
    6x=0
    x=0

    fy=4y-4
    4y-4=0
    4y=4
    y=1

    Critical point at (0,1)

    and I know how to test for extrema at this point
    fxx = 6
    fyy = 4
    fxy = 0

    d= 6*4 - 0 = 24
    d>0 so there is a saddle point at (0,1,-2)

    I do not, however, know how to test for extrema at the bounds. I know I need to be looking at f(x,4) and f(x,x2) but i don't know what to do with them. If anyone could give me directions I would greatly appreciate it
     
    Last edited: Dec 8, 2013
  2. jcsd
  3. Dec 8, 2013 #2

    Simon Bridge

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    you know how to find extreema in y=f(x) don't you?
     
  4. Dec 8, 2013 #3
    ok so i think I've made some progress but I'm still unsure if I'm doing it correctly.
    for y=4
    f(x,4) = 3x2+16
    d/dx = 6x
    6x = 0 at x = 0
    so (0,4) ?
    f(0,4) = 16
    this would end up being my absolute maximum
    for y=x2
    f(x,x2) = 2x4-x2
    d/dx = 8x3-2x
    8x3-2x = 0
    2x (4x2-1)=0
    4x2-1=0
    4x2=1
    x2=1/4
    x=√(1/4)=1/2
    then plug x=1/2 back into the original function
    3(1/2)2+2y2-4y=0
    -3/4=2y2-4y
    divide both sides by 2 and complete the square
    5/8=(y-1)2
    √(5/8) = y-1
    y=(√(10)/4) +1
    so f(1/2, (√(10)/4) +1) = 3/4 which ends up being my absolute minimum

    can anyone tell me if this is right?
     
  5. Dec 8, 2013 #4

    Ray Vickson

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    You have mis-characterized the nature of the point (0,1).

    Have you taken the Karush-Kuhn-Tucker conditions yet? If so, use them to characterize the boundary extrema (local max or saddle point). If you have not taken them yet, just look at the relationship between the gradient vector of f and the tangent directions at the tight constraints: (1) on a single constraint boundary, the gradient must point out of the feasible region and be perpendicular to the tangent line of the constraint at that point; and (2) at an intersection of two constraint boundaries the gradient of f must point outward and be between the two perpendiculars to the two tangent lines of the constraint boundaries (draw yourself a picture to clarify).
     
  6. Dec 8, 2013 #5

    HallsofIvy

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    First, there is no point in finding the second partials. That test is to determine whether or not a critical point is a relative maximum or minimum and you are asked for absolute extrema.

    Any possible extremum in the interior of the region must be where the partial derivatives are 0 so first find where [itex]f_x(x, y)= 0[/itex] and [itex]f_y(x,y)= 0[/itex] and list them.

    Now look on the boundaries. One boundary is y= 4 and [itex]f(x, 4)= 3x^2- 8[/itex]. Find any points where the derivative of that is 0 and add them to the list.

    Another boundary is [itex]y= x^2[/itex] and [itex]f(x, x^2)= 2x^4- x^2[/itex]. Find any points where the derivative of that is 0 and add them to the list.

    Finally, look at the "boundaries of the boundaries". That is, the points (2, 4) and (-2, 4) where the line y= 4 and the curve [itex]y= x^2[/itex] intersect. Add those to the list.

    The only way to determine absolute extrema is to actually calculate f(x, y) for all of the points now on that list (I get 6) and observe which value is smallest and which is largest.
     
    Last edited: Dec 8, 2013
  7. Dec 8, 2013 #6

    Ray Vickson

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    Actually, we can say more in this case, but using material that is, perhaps, beyond what the OP has seen. In this case the function ##f(x,y)= 3 x^2+2y^2-4y## is strictly convex, so any stationary point is automatically a global min in the whole plane. If that stationary point lies inside or on the boundary of the feasible region, it is the min in the constrained problem as well. Furthermore, since the constraint region is a convex set and f is strictly convex, the constrained max lies at an extreme point of the feasible region, and these are the two upper corners plus the lower parabola; points on the upper boundary line strictly between the corners are not candidates for the max.

    However, as I indicated, these are theorems, etc., in optimization courses and may be unavailable to the OP.
     
  8. Dec 8, 2013 #7

    vela

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    I assume you concluded this after calculating f(x,y) at the other points.

    That should be f(1/2, 1/4) because you're trying to find the extrema of f(x, x2).

    What happened to the root x=0?
     
  9. Dec 8, 2013 #8

    Ray Vickson

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    Nope: the function f can take values < 0 in R, so your alleged "minimum" value of +3/4 is not at all a minimum.
     
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