Critical points and 2nd partials test

camino
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Homework Statement



Find all critical points of f(x,y) = 12xy-x^2 y-2xy^2 and test them for relative extrema using the 2nd partials test.

Homework Equations





The Attempt at a Solution



{0=fx(x,y)=12-2x-2 10x-2 x=5
{0=fy(x,y)=12-4y y=3

Critical point: (5,3)

fxx(x,y)= -2
fyy(x,y)= -4
fxy(x,y)= 0

d(5,3)=fxx(5,3)fyy(5,3)-(fxy(5,3))^2
d=(0)(0)-(0)
d=0 no conclusion
---------------------------------------

Just wondering if I did this right?
 
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camino said:

Homework Statement



Find all critical points of f(x,y) = 12xy-x^2 y-2xy^2 and test them for relative extrema using the 2nd partials test.

Homework Equations





The Attempt at a Solution



{0=fx(x,y)=12-2x-2 10x-2 x=5
{0=fy(x,y)=12-4y y=3

Critical point: (5,3)

fxx(x,y)= -2
fyy(x,y)= -4
fxy(x,y)= 0

d(5,3)=fxx(5,3)fyy(5,3)-(fxy(5,3))^2
d=(0)(0)-(0)
d=0 no conclusion
---------------------------------------

Just wondering if I did this right?
fxx(5,3)fyy(5,3)= (-2)(-4), not (0)(0)!
 
Ok, so

d=(-2)(-4)-(0)
d=8

Is that right?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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