absolute max/min

2013-06-02 11:56 am
Find the absolute maximum and minimum of the function f(x,y) = y sqrt(x) - y^2 - x + 3y on the domain 0=<x=<9, 0=<y=<4.
absolute minimum value = __ attained at ____.
absolute maximum value = ___ attained at ___.

my attempt:
f_x = 0.5 y x^(-0.5) - 1
f_y = sqrt(x) - 2y + 3

then I set the two partial derivatives above to be zero,
so I found a critical point (1,2).

then I do:
D(1,2) = 0.25
and 0.25 is > 0, so it's some kind of extremum.
and therefore I have to check f_xx(1,2), which is equal to -0.5.
and -0.5 <0, so it's a local max.

then I have to check the boundary.
and I found f(0,0) = 0,
f(0,4) = -4
f(9,4) = -1
f(9,0) = -81.
so I know that absolute minimum value is -81 at point (9,0).
but I believe there's some more points that are also absolute minimum. and some more points that are absolute maximum.

please help. thanks!

回答 (2)

2013-06-03 12:14 am
✔ 最佳答案
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圖片參考:http://imgcld.yimg.com/8/n/HA05107138/o/20130602161354.jpg
2013-06-02 10:04 pm
you should parametrize the "edge" of the boundary , namely
the straight line joining any two points of the (0,0),(0,4),(9,0),(9,4).
And substitute back into f(x,y) to see whether it is maximum or minimum.

2013-06-02 14:06:00 補充:
PS : after substituting back into f(x,y) , you can determine its max / min by taking
derivatives .


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