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Absolute extrema 2 variable function

  1. Oct 24, 2015 #1
    1. The problem statement, all variables and given/known data
    find the absolute extrema of f(x,y) = 2x - 2xy + y^2
    in the region in the xy plane bounded by the graphs of y= x^2 and y = 1

    3. The attempt at a solution
    first we find the first partials
    fx(x,y) = 2 - 2y
    fy(x,y) = 2y-2x
    2-2y = 0 when y = 1
    2y - 2x = 0 when y=x in this case y=1 so x =1
    f(1,1) = 1

    now we check along y = 1
    f(x,1) = 2x - 2x + 1 = 1
    so z = 1 along the entire line y =1 . this includes the points where y=1 intersects with y=x^2

    now we check along y = = x^2
    f(x, x^2) = 2x -2x^3 + x^4
    f'(x,x^2) = 2 - 6x^2 + 4x^3
    we need this to equal 0
    2 - 6x^2 + 4x^3 = 0
    1 - 3x^2 + 2x^3 = 0
    now I am a little lost. I need to solve this for x but there doesn't seem to be an easy way to do this. After I solve for x I plug the corresponding point into the original function to find the minimum. (the minimum has to be along this curve because it was nowhere else and it is a bounded region) I could be doing this wrong though as these types of problems are very confusing to me .
     
  2. jcsd
  3. Oct 24, 2015 #2

    Ray Vickson

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    Homework Helper

    If you want the max or min of f(x,y) along the parabola, just solve the equality-constrained problem
    [tex] \begin{array}{rl}\max / \min & 2x - 2 xy + y^2\\
    \text{subject to} &y - x^2 = 0
    \end{array}
    [/tex]
    You can do that fairly easily using the Lagrange multiplier method.

    Alternatively, you can try to solve your equation ##2 x^3 - 3 x^2 + 1 = 0##, using the Rational Root Theorem, for example. Or, you can do it numerically if you are satisfied with an approximate numerical answer.
     
  4. Oct 24, 2015 #3
    Ok i will use lagrange multipliers.
    The gradient of f=(2-2y)i + (2y-2x)j
    Let g= y-x^2 =0
    Then the gradient of g
    = -2xi + j
    Multiply the gradient of g by a constant c
    = -2cxi + cj

    We must solve the system
    2-2y = -2cx
    2y-2x = c
    Y-x^2 = 0
    Plug in our value for c into the first equation
    2-2y=-2(2y-2x)x
    2-2y = -4yx + 4x^2
    Plug in y=x^2
    2-2x^2= -4x^3 + 4x^2
    2= 6x^3 - 4x^3

    Wait i ended up with the same equation. I cant figure out where i went wrong
     
  5. Oct 24, 2015 #4
    Thats supposed to be a 6x^2
     
  6. Oct 24, 2015 #5
    I think I figured it out using the rational roots theorem.
    for the equation 2x^3-3x^2 + 1
    the possible rational roots are
    -1/2 , 1/2 , -1 and 1
    plugging them all in the only ones that give 0 are
    1 and -1/2
    since y = x^2 if x =1 y = 1 we already know z = 1 at this point
    if x = -1/2 y = 1/4
    plugging this into the original equation
    f(-1/2, 1/4) = 2(-1/2) - 2(-1/2)(1/4) + (1/16)
    = - 1 + (1/4) + (1/16)
    = -(16/16) + (4/16) + (1/16)
    = (5-16)/16
    = -11/16
    so the min is z = -11/16
    and the max is z = 1
     
  7. Oct 24, 2015 #6

    Ray Vickson

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    Science Advisor
    Homework Helper

    Use the equations to solve for x and y in terms of c, giving
    [tex] x = x(c) = \frac{2-c}{2(1-c)}, \; y = y(c) = \frac{2-c^2}{2(1-c)} [/tex]
    Then plug those expressions into the function g, which will now be a function ##g = G(c)## of c alone. You set ##G(c) = 0## to figure out the value (or values) of c. In this case,
    [tex] g(x(c),y(c)) = G(c) = -\frac{c^2 (3 + 2c)}{4(1+c)^2}, [/tex]
    so we need (i) ##c = 0##, giving ##x=1, \:y=1##; or (ii) ##c = -3/2##, giving ##x=-1/2, \:y = 1/4##.

    That is usually (not always) the way these things are done: express the variables in terms of the multiplier, then use the constraint to fix the value of the multiplier.
     
  8. Oct 24, 2015 #7
    thank you
     
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