critical points of a multivariable function

[erinhannen]

The problem asks to find the absolute maximum and minimum values of f on the set D.

f(x,y)=(x^3)-(3x)-(y^3)+(12y), D is the quadrilateral whose vertices are (-2,3), (2,3),(2,2),(-2,-2).

I understand how to go about the problem, just have a small question. I started by taking the first partial derivatives with respect to x and y so I get fx=3x^2-3 and fy=-3y^2+12. Setting each of those equal to 0 you can solve x=1,-1 and y=4, -4, but what are the critical points from these values? Are they (1,4), (1,-4), (-1,4), and (-1,-4)? Or is it just (1,4) and (-1,-4)...? From there I know I need to find if there are relative extrema on the boundary, but I'm just not understanding if all 4 of those are critical values...? Thanks for the help!

 

[galactus]

\(\displaystyle f(x,y)=x^{3}-3x-y^{3}+12y\).........[1]

\(\displaystyle \frac{{\partial}f}{{\partial}x}=3x^{2}-3=0\Rightarrow x=\pm 1\)

\(\displaystyle \frac{{\partial}f}{{\partial}y}=-3y^{2}+12=0\Rightarrow y=\pm 2\)

2 of these critical points lie inside the region. (-1,2), (1,2).

The boundary of the region consists of four line segments. Each treated separately.

The line between (-2,3) and (2,3). On this segment we have y=3, so [1] simplifies to \(\displaystyle x^{3}-3x+9\)

\(\displaystyle \frac{d}{dx}[x^{3}-3x+9]=3x^{2}-3\). See there?. The same as above when we took the partial w.r.t x.

Continue on an test the other three line segments. Make a chart.

 

[erinhannen]

thanks so much!