gradient of F = (3x^2-3, -2y+6 ) =(0,0),
if 3x^2-3=0 and -2y+6=0
so
x^2=1 and y=3
(x=1 or x=-1) and y=3
so there are 2 points:
(1,3) and (-1,3)
f_xx = 6x
f_xx f_yy -(f_xy)^2 = 6x(-2) -(0 )^2=-12x
point (1,3):
f_xx (1,3)f_yy(1,3) -(f_xy(1,3))^2 = -12(1)<0 so (1,3) is a saddle point.
point (-1,3):
f_xx(-1,3)=6(-1)<0
and
f_xx (-1,3)f_yy(-1,3) -(f_xy(-1,3))^2 = -12(-1) >0, so (-1,3) is a local maximum
2006-10-25 03:16:36
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answer #1
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answered by Anonymous
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I onl;y tell yu the steps.
1) the critical points are solutions of dF(x,y) = 0
dF(x,y) is a matrix 2x2
..... look in your book this textbox is not graphic enough to show you the solution...
2006-10-23 20:10:47
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answer #2
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answered by gjmb1960 7
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