Finding absolute minimum and maximum values

[grog]

Homework Statement

Find the absolute minimum and maximum values of f on the set D.

f(x,y)= e^-x2-y2(x²+2y²); D is the disk x²+y² <= 4

Homework Equations

Second Derivatives test,
partial derivatives

The Attempt at a Solution

f_x(x,y) = 0 = (e^-x2-y2)(-2x) + (x²+2y²)(-2x e^-x2-y2)

f_y(x,y) = 0 = (e^-x2-y2)(4y) + (x²+2y²)(-2y e^-x2-y2)

f_xy(x,y) = (e^-x2-y2)+(-2x)(-2y e^-x2-y2) + (x²+2y²)(-2x*-2y e^-x2-y2) + (-2x e^-x2-y2)(4y)

f_x and f_y simplify to:
f_x (x,y) = 1+x²+2y² = 0
f_y (x,y) = -2y+x²+2y² = 0

I'm stymied here because the equation I get for f_x seems impossible to solve. Did I make a mistake differentiating?

 

[Dick]

Yes, there's a sign error in f_x. But f_y looks ok, and I don't think (4y)-(2y)*(x^2+2y^2) simplifies to what you got for f_y.

 

[grog]

ah.

ok.

so I should have ended up with
f_x= x²+2y² -1 = 0 and
f_y=x²+2y²-2 = 0

So to find the critical points I have to solve these two equations.

Would I be on the right track to say f_xthen simplifies to (x-1)(x+1)+2y² = 0


If so, I'm embarassed to say I'm still not sure how to proceed from here. I find the examples in my book don't really adapt themselves very well to the exercises, so I'm having trouble getting to the next step. : (

 

[grog]

ok, so it looks like I got my approach wrong.

The way it's worked out it looks like the e distributes to the other term, and then I should differentiate, which yields


e^{-x^2-y^2}*2x =- 2x^3*e^{-x^2-y^2}

and
e^{-x^2-y^2} * 4y-4y^3e^{-x^2-y^2}


which reduces to
2x(1-x^2) = 0
4y(1-y^2) = 0

so we have critical points at x=0,1,-1
and cp at y=0,1,-1

we also need to check for extreme values on the boundary.
now, the boundary function is $$x^2+y^2=4$$ so $$e^(-x^2+y^2)$$ is $$e^-4$$

and means that $$x^2=4-y^2$$
so if we plug into our original function, we get

$$f(y)=e^-4 (4-y^2+2y^2)$$ for -2<=y<=2

so
$$f'(y)=e^-4*2y=0$$
solving for y we get y=0, and plugging that into our boundary equation $$x^2+y^2=4$$ and solving for x we get x=+/- 2

There's more, where I have to set something up with Lagrange, but I'm still working on understanding that part. I'll post the rest once I understand it a little better.

 

[HallsofIvy]

Of course, it doesn't necessarily follow that there is a critical point inside the circle- and even if there is the max or min are not necessarily there. It is also possible for the max or min (or both) to occur on the boundary of the set which, here, is the circle itself.

 

[Dick]

I get f_x=2x-2x(2y^2+x^2)*exp(...) and f_y=4y-2y(2y^2+x^2)*exp(...) and you came pretty close to getting it right the first time. I don't know what you did in your last response. I get five critical points.