Local Extrema with Partial d/dx

[Oglethorpe]

Hello, I'm been stuck on this problem and I've been staring blankly at it way too long. I stumbled upon here and thought I'd ask for help? :P

Alright well, I'm looking for a local max/min, and I've already done the first partials and I got *f(x)=2x-y and f(y)=-x+2y+6; I'm sure those are right and the seconds partials were; f(xx)=2, f(xy)=-1, and f(yy)=2

...well, how do I use the Second derivative test on that? Where do I plug my critical point in (which I got (3/2, -6) is that right?)? Anyways, help would be great! :D

*It really means f subscript x, and so do the others, I just don't know how to use subscripts on a keyboard.

 

[Dick]

Just use '_' for subscript, as in f_x=2x-y. (3/2,-6) doesn't satisfy f_x=0, does it? Why do you think that would be a critical point?

 

[Oglethorpe]

Oh thanks, and I simplified the f_x to y=2x-9, which I then substituted into f_y and went from f(y)=-x+2y+6 to -2x+9+4x-18+6=0, which simplifies to 2x-3=0 resulting in x=3/2. I'm more curious on how to take a second derivatives test when there are no variables to plug in a critical point, however.

 

[Dick]

Oh thanks, and I simplified the f_x to y=2x-9, which I then substituted into f_y and went from f(y)=-x+2y+6 to -2x+9+4x-18+6=0, which simplifies to 2x-3=0 resulting in x=3/2. I'm more curious on how to take a second derivatives test when there are no variables to plug in a critical point, however.

Well, x=3/2 is still wrong for a critical point, but ok, the second derivative test just gives you a constant, yes? Is it positive or negative? There's no need to plug anything in in this case.

 

[Oglethorpe]

Why would that be? And yes, I get

f_xx=2
f_xy=-1
f_yy=2

 

[Dick]

Why would that be? And yes, I get

f_xx=2
f_xy=-1
f_yy=2

x=3/2 is wrong as a critical point. If you can tell me what you think the corresponding y is then it should be obvious. Unless you mistyped f_y or f_x.

 

[Oglethorpe]

Oh! I typed f_x=2x-y when I should've typed f_x=2x-y-9

 

[HallsofIvy]

The point is that your second derivatives are constant functions. You do not put in the values of x and y for the critical point anywhere. $f_xxf_yy- fxy^2$ is just (2)(2)- 1= 3> 0.

When the second derivatives are constants, the first derivatives must be linear so there can be only one critical point anyway.