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Math Help - Maximum/minimum values

  1. #1
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    Maximum/minimum values

    Find the local max/min values and saddle points:

    f(x,y) = (1+xy)(x+y)

    taking the partial derivatives I get

    f_x = y(x+y) + (1+xy) \ \ \mbox{and} \ \ f_y = x(x+y)+(1+xy)

    f_x = y(x+y) + (1+xy) = 0

    y(x+y) =  - (1+xy)

    xy+y^2 =  -1-xy

    x  =  -\frac{y}{2} - \frac{1}{2y}

    taking:

    f_y = x(x+y)+(1+xy) = 0

    y = -\frac{x}{2} - \frac{1}{2x}

    if I make the replacement I get:

     y= \left( \frac{y}{4} - \frac{1}{4y} - \left( \frac{1}{-y-\frac{1}y} \right) \right) = \frac{5y}{4} + \frac{3}{4y}

    is this correct?
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  2. #2
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    Quote Originally Posted by lllll View Post
    Find the local max/min values and saddle points:

    f(x,y) = (1+xy)(x+y)

    taking the partial derivatives I get

    f_x = y(x+y) + (1+xy) \ \ \mbox{and} \ \ f_y = x(x+y)+(1+xy)

    f_x = y(x+y) + (1+xy) = 0

    y(x+y) =  - (1+xy)

    xy+y^2 =  -1-xy

    x  =  -\frac{y}{2} - \frac{1}{2y}

    taking:

    f_y = x(x+y)+(1+xy) = 0

    y = -\frac{x}{2} - \frac{1}{2x}

    if I make the replacement I get:

     y= \left( \frac{y}{4} - \frac{1}{4y} - \left( \frac{1}{-y-\frac{1}y} \right) \right) = \frac{5y}{4} + \frac{3}{4y}

    is this correct?
    You're hoeing the hard road.

    \frac{\partial f}{\partial x} = y^2 + 2xy + 1 = 0 .... (1)

    \frac{\partial f}{\partial y} = x^2 + 2xy + 1 = 0 .... (2)

    (1) - (2): y^2 - x^2 = 0 \Rightarrow (y - x)(y + x) = 0 \Rightarrow y = \pm x.

    Case 1: y = x

    Substitute into (1): 3x^2 + 1 = 0. No real solution.

    Case 2: y = -x

    Substitute into (1): -x^2 + 1 = 0 \Rightarrow x = \pm 1.


    Extremum solutions are (1, -1) and (-1, 1).

    Now you have to test their nature.
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