Locating relative extrema and saddle points.

[Treize]

I have:
f(x,y) = 3x[sup:20rvm84k]2[/sup:20rvm84k] - 2xy + y[sup:20rvm84k]2[/sup:20rvm84k] - 8y

f[sub:20rvm84k]x[/sub:20rvm84k](x,y) = 6x - 2y = 0
f[sub:20rvm84k]y[/sub:20rvm84k](x,y) = -2x + 2y - 8 = 0

solving for x and y I get x = 2, y = 6, (2,6) = only crit pt.

Now, this is where I'm having a bit of a problem, applying the second partials test.
I can get f[sub:20rvm84k]xx[/sub:20rvm84k] and f[sub:20rvm84k]yy[/sub:20rvm84k] no problem, but I can't remember how to find f[sub:20rvm84k]xy[/sub:20rvm84k]. Could someone point out how to obtain it?

 

[DrMike]

I have:
f(x,y) = 3x[sup:3rrudg92]2[/sup:3rrudg92] - 2xy + y[sup:3rrudg92]2[/sup:3rrudg92] - 8y

f[sub:3rrudg92]x[/sub:3rrudg92](x,y) = 6x - 2y = 0
f[sub:3rrudg92]y[/sub:3rrudg92](x,y) = -2x + 2y - 8 = 0

solving for x and y I get x = 2, y = 6, (2,6) = only crit pt.

Now, this is where I'm having a bit of a problem, applying the second partials test.
I can get f[sub:3rrudg92]xx[/sub:3rrudg92] and f[sub:3rrudg92]yy[/sub:3rrudg92] no problem, but I can't remember how to find f[sub:3rrudg92]xy[/sub:3rrudg92]. Could someone point out how to obtain it?

I'll do a different example. Suppose
\(\displaystyle g(x,y) = x^3y-sin(x+4y)\)
Then :
\(\displaystyle \frac{\partial}{\partial x}g(x,y) = 3x^2y-cos(x+4y)\)
so that :
\(\displaystyle \frac{\partial^2}{\partial x\partial y}g(x,y) = \frac{\partial}{\partial y}3x^2y-cos(x+4y) = 3x^2+4sin(x+4y)\)
You could also work this out as
\(\displaystyle \frac{\partial^2}{\partial x\partial y}g(x,y) = \frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}g(x,y)\right)\).