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Absolute Max & Absolute Min

  1. Mar 5, 2006 #1
    I have been understanding these so far but this one it says is wrong, but I cant figure out why.

    "Find the absolute maximum and absolute minimum of the function f(x,y) = xy - 5 y - 25 x + 125 on the region on or above y = x^2 and on or below y = 29."

    I get the absolute max to be =148.15 at (-5/3,25/9), which is correct.
    I get the absolute min to be =0 at (5,25), which is wrong, but I cant find any more points of interest, as far as I can see 0 at (5,25) is right :grumpy:

    If you have any clues/suggestions, I would greatly appreciate it :smile:
  2. jcsd
  3. Mar 5, 2006 #2
    You have not properly examined what happens to f on the boundary of the region.
  4. Mar 5, 2006 #3
    which boundry? y=29 or y=x^2?

    for y=x^2 i get the points (5,25) and (-5/3,25/9)
    for y=29 i get no points

    I have looked over each one a half dozen times, if you know which one the problem is in, id like to know thx.
    Last edited: Mar 5, 2006
  5. Mar 5, 2006 #4
    Well, I know there's at least one point on "the y = 29 boundary" where f is negative. When you consider what happens to f restricted to y = 29, you are checking the endpoints of the interval (which x can range over) and not just the points where the derivative equals 0, right?
  6. Mar 5, 2006 #5
    ahhh, because at the endpoints it may be an extrema yet the derivative is non-zero. Got it! Thx alot!
  7. Mar 5, 2006 #6
    so if my range is a circle, there are no endpoints to check right?
  8. Mar 5, 2006 #7


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    But the entire circle would be the boundary. Write the boundary in parametric equations, plug those parametric equations into the object function to get a function of one variable and then differentiate that.

    In this example, where the boundary was the parabola y= x2 and y= 29, you would need to look for max and min of
    f(x, 29)= 29x- 5(29)- 25x+ 125= 4x- 145+ 125= 4x- 20 for the y= 29 part of the boundary. That never has derivative 0 so you look for max and min at its endpoints, [itex]x= \pm \sqrt{29}[/itex].

    f(x,x2= x3- 5x2- 25x+ 125 has derivative 3x2- 10x- 25 which is never 0 and so you look for max and min at its endpoints, also [itex]x= \pm \sqrt{29}[/itex].
    Last edited by a moderator: Mar 5, 2006
  9. Mar 5, 2006 #8
    ugg, sorry, but now Im even more confused on the next one.

    "Find the absolute maximum and absolute minimum of the function f(x,y) = 2x^3 + y^4 on the region {(x,y) | x^2 + y^2 =< 16} "

    When writing the boundry in parametrics, i use x=cos(theta) y=sin(theta)? THis gives me 1=<16 for my boundry...
  10. Mar 6, 2006 #9


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    You have the wrong parametric equations. The boundary is the circle
    x2+ y2= 16 (the interior of the circle, < 16, is not the boundary!). The parametric equations are x= 4 cos([itex]\theta[/itex]), y= 4 sin([itex]\theta[/itex]).

    And you don't put those into the equation for the boundary (which would give you the trivial 16= 16), you put them into the equation for f:
    f(x,y)= f([itex]\theta[/itex])= 2(4)3cos3([itex]\theta[/itex])+ 44sin4([itex]\theta[/itex]).
  11. Mar 6, 2006 #10
    Thx for all your help, I really do appreciate it.

    I have been trying and trying this but Im simply unable to do it. I have the derivative of the boundry function in terms of theta, but I do not know how to solve it and find what values of theta give me 0. Im gonna go to the library and try to find a trig. book.
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