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Math Help - Tough Max/Min Problem

  1. #1
    Member Em Yeu Anh's Avatar
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    Question Tough Max/Min Problem

    Let  F(x,y) = 2(1-x)^{\frac{1}{2}}+(x+y)^2+x-2y be a function that is bounded by x=-3, y(t)=t^2, x(t)=2t+t^2
    I am to find the absolute max/min values.
    Completely lost! This is significantly more difficult that any examples we went through in class.
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  2. #2
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    You appeared to have copied the problem wrong. x= t^2+ 2t= t^2+ 2t+ 1- 1= (t+1)^2- 1. No matter what t is, x is never less than -1 and, in particular, is never equal to -3. Since the "boudaries" given do not intersect, they cannot bound a region.
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  3. #3
    Member Em Yeu Anh's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    You appeared to have copied the problem wrong. x= t^2+ 2t= t^2+ 2t+ 1- 1= (t+1)^2- 1. No matter what t is, x is never less than -1 and, in particular, is never equal to -3. Since the "boudaries" given do not intersect, they cannot bound a region.
    Sorry, should be x=t^2-2t.
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  4. #4
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    Mathematics requires great precision. If you do not take the time and care to read the problem correctly you are are never going to get anywhere. " x= t^2- 2t" has exactly the same difficulty as " x= t^2+ 2t". x= t^2- 2t= t^2- 2t+ 1- 1= (t- 1)^2- 1 so that x is never less than -1. The curve given by x= t^2- 2t, y= t^2 does not intersect x= -3 so the curves you give do NOT bound a region.

    If it were x= 2t- t^2= 1- (1- 2t+ t^2)= 1- (t- 1)^2, x= -3 when -3= 1- (t-1)^2 or (t-1)^2= 4, t- 1= \pm 2, t= -1 and t= 3. At those points, y= 1 and 9.

    Now, find the gradient of F(x,y) and set it equal to 0 to find any points in the interior of the region at which there may be max or min.

    Then set x=-3 in F(x,y) to get a function in the single variable y giving the value of F on the line x= -3. You can set the derivative of that equal to 0 to find any critical points there (remember that y must be between 1 and 9 to be on that boundary).

    Set x= 2t- t^2, y= t^2 in F to get a function of the single variable t on the parabola. Set its derivative equal to 0 to find any critical points there (remember that t must be between -1 and 3).

    Evaluate the function F(x,y) at each of the critical points you have found, as well as at the vertices (-3, 1) and (-3, 9) to determine the largest and smallest values of F in that region.
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