Find the extreme values of $f(x,y)=x^2+y^2+4x-4y$ on the region described by $x^2+y^2<=9$

Question

I've got

$2x+4=2x\lambda$ , $2y-4=2y\lambda$

$\lambda =\frac{x+2}{x}$, $\lambda=\frac{y-2}{y}$

$\frac{x+2}{x}=\frac{y-2}{y}$

$(\frac{x+2}{x})^2+(\frac{y-2}{y})^2$ = 9 //based on another that just used equals the right hand side number like this

and $ x^2+4x+y^2-4y=1$ which didn't get me anywhere.

Also tried

$\frac{y-2}{y}=\frac{x+2}{x}=1+\frac{2}{x}$

$\frac{-2}{y}=\frac{2}{x}$

$(\frac{-2}{y})^2+(\frac{2}{x})^2=9$

The answer are min at (-2, 2) and max at $(\frac{3}{ \sqrt{2}} \frac{-3}{ \sqrt{2}})$

I can kinda of see where you could get the first point from the second thing I tried but no clue about the second point.

Answer 1

Just because you have found that

$\frac{x+2}{x}=\frac{y-2}{y}$ doesn't mean that $x = \frac{x+2}{x}$ and you can plug it into your equation $x^2 + y^2 = 9$

$x = \frac {-2}{1-\lambda}, y = \frac {2}{1-\lambda}$

And plug that into:

$x^2 + y^2 \le 9\\ (-2)^2 + 2^2 \le 9 (1-\lambda)^2\\ 0\ge 1-\lambda\le \frac {\sqrt {8}}{3}$

Now before applying calculus, it should be clear: $f(x) = (x+2)^2 + (y-2)^2 - 8$

$f(x)$ is the square of the distance from the point (-2,2) less a constant.

The absolute minimum is at $(-2,2)$

The maximal distance from the point, inside the disk, is on the boundary of the disk, straight across from the center of the disk.

Answer 2

Hint. Note that $$f(x,y)=x^2+y^2+4x-4y=(x+2)^2+(y-2)^2-8\geq -8=f(-2,2)$$ Since $(-2,2)$ satisfies $x^2+y^2\leq 9$, it is the global minimum.

Now the maximum should stay on the boundary of the region, that is the circle $x^2+y^2=9$ (why)?, where $f(x,y)=9+4x-4y$.

Answer 3

First candidate for extrema, momentarily ignoring the constraint, where both partials are zero $$ 2x+4=0,2y-4=0\Rightarrow (x,y)=(2,-2) $$ Which is inside the region, as I think you found. As this is a compact space, you must also check the boundary, or where $$ x^2+y^2=9 $$ But maximizing this is just a normal lagrangian with constraint that x and y lie on the circle.