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Extrema problem

  1. Jul 17, 2007 #1
    1. The problem statement, all variables and given/known data

    Find and classify the local and absolute extrema of the function
    [tex]f(x,y)=x-xy[/tex] over the region
    [tex]D={(x,y)|x^2+y^2\leq1 and x+y\geq0}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    Critical points are where the first derivative (gradient) is 0.
    [tex]Grad(f)=(1-y, -x)=0[/tex]
    So critical point a=(0,1)

    In order to classify the critical point, find the Hessian matrix of f at a:
    Then the quadratic form is:

    Which means the test is inconclusive??? I.e., the critical point a is a saddle point, which is neither local maximum or minimum. Am I right?

    Also, how do you find the absolute extreme of the function on the region aforementioned?

    I tried to convert x^2+y^2<=1 to polar coordinates, which gives
    However, this doesn't help much as the original function converted to polar coordinates is not straightforward to find its extrema given the domain of r and theta.
  2. jcsd
  3. Jul 17, 2007 #2


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    You've shown there are no critical points in the interior. So now analyze the boundary. Parametrize the boundary arc and segment and look at the function along them. The extrema must be there.
  4. Jul 17, 2007 #3
    How do you parameterize f(x,y)=x-xy?

    One thing I tried and was mentioned in my OP is polarize x,y, since [tex]x^2+y^2\leq1[/tex], therefore, let [tex]x=r\cos\theta[/tex], [tex]y=r\sin\theta[/tex], then [tex]0<r\leq1[/tex], and from [tex]x+y\geq0[/tex], we can fix the domain for [tex]\theta[/tex] as well.

    However, then [tex]f(x,y)->f(r,\theta)=r\cos\theta-r\cos\theta r\sin\theta[/tex], and applying trigonometric identities doesn't lead me anywhere to solve the range for [tex]f(r, \theta)[/tex]
  5. Jul 17, 2007 #4


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    Parametrize the boundary. One part of the boundary is an arc of radius 1 from theta=-pi/4 to 3pi/4. I would parametrize that as x=cos(theta), y=sin(theta). Put that into x-xy and get a function only of theta. The other part of the boundary is a line segment connecting the arc endpoints. How would you parametrize that?
  6. Jul 17, 2007 #5
    You mean parameterize a line segment? Since the line goes through (-1,1) and (1,-1), then [tex]f(t)=(1+2t, 1-2t)[/tex], so transform f(x,y) to f(t):
    [tex](1+2t)(1-(1-2t))=2t(2t+1)=4t^2+2t=4(t+1/4)^2-1/4[/tex], so f(t) has min value -1/4 at t=-1/4, right?

    What about the other part? [tex]f(\theta)=\cos\theta-\cos\theta \sin\theta[/tex], given the domain of theta, it's still hard to find the extrema for this function since [tex]\cos\theta \sin\theta[/tex] is bound to get a trig function with [tex]2\theta[/tex] and f is going to be like [tex]\cos\theta - something(2\theta)[/tex], which is hard to find a bound on.
  7. Jul 17, 2007 #6


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    Correction, the endpoints are like (-sqrt(2)/2,sqrt(2)/2), but you have the right idea. Don't forget to check for extrema AT the endpoints. As for the trig stuff, don't complain before you try it. You just might be able to turn it into a quadratic equation in sin(theta) rather than going the double angle route. Hint, hint.
  8. Jul 17, 2007 #7

    Well, let's see...
    [tex]\cos\theta \sin\theta
    =\frac{1}{2}\sqrt{1-\cos^2 2\theta}
    =\cos\theta \sin\theta[/tex]
  9. Jul 17, 2007 #8


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    f'(theta)=-sin(theta)-cos^2(theta)+sin^2(theta). Unless I've fluffed a sign. cos^2(theta)=1-sin^2(theta), right? See? It's quadratic.
  10. Jul 18, 2007 #9
    This may not help at all, but doesnt that function represent a family of rectangular hyperbolas with the axis at (0,1)? Perhaps you could use that to parameterize the function. The parametric point on a hyperbola is (asec(theta), 1-btan(theta)).

    So what you really want to find is the maximum possible value of that function above the line x+y=0, and inside the unit circle x^2+y^2=1 right?

    Since this is a rectangular hyperbola, the max value should be where the hyperbola cuts the circle in the first quadrant. Maybe you could find the general points of intersection of the two curves. Dunno if that helps or not.
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