Find the extrema of f.
f(x,y)=x-y, subject to x^2-y^2=2

Question

Find the extrema of f.
f(x,y)=x-y, subject to x^2-y^2=2

tkhunny · Answer

How shall we do that?  Lagrange or some other way?

anonymous · Answer

Lagrange would be best.

tkhunny · Answer

Well, set up your giant, new function: $f(x,y,\lambda) = x - y - \lambda(x^{2} - y^{2} - 2)$.

Now find the three partial derivatives.  Do you need help with that?

anonymous · Answer

I'm getting (0,0,-1).

tkhunny · Answer

Waht were your partial derivatives?

anonymous · Answer

Sorry I messed up. I got $$1-\lambda 2x=0, -1-\lambda 2y=0, x^2-y^2-2=0$$

tkhunny · Answer

Unfortunately, that's not all that useful.  We probably should notice that the contraint can be rewitten $x - y = \dfrac{2}{x+y}$ except on x = -y.

anonymous · Answer

$$
f(x,y)=x-y\
g(x,y)=x^2+y^2\
f_x(y)=1\
f_y(x)=-1\
g_x(y)=2x\
g_y(x)=2y\
f_x=\lambda g_x\
2x=\frac1\lambda\
f_y=\lambda g_y\
-2y=\frac1\lambda\
x=-y\
g(x,y)=2\
2y^2=2\
(1,-1,2), (-1,1,-2)\
$$