Find the absolute maximum and minimum of $f(x,y)=exp(x^2-y^2)$ on the set $R^2$ with $x^2+y^2 \leq1$.
I have applied Lagrange Multipliers method, but found that exponent function should be equal to zero if $f$ function has minimum. Any help will be strongly appreciated.
On
Changing to polar coordinates
$$ \cases{ x=r\cos\theta\\ y=r\sin\theta } $$
we get the equivalent problem
$$ \min(\max)e^{r^2\cos(2\theta)}\ \ \ \text{s. t.}\ \ \ 0\le r\le 1 $$
now along a direction $\theta_0$ as $\{-\theta_0 r\cup\theta_0 r\}$ for $0\le r\le 1$ : for $\theta \ne \pm\frac{\pi}{4}+k\pi,\ k=0,1,2,\cdots\ $ we have curves in which $\cos(2\theta_0)> 0$ and also curves in which $\cos(2\theta_0) < 0$. For the first curves we have a minimum at $r = 0$ and for the second we have curves with a maximum at $r=0$. Those curves are separated at $\theta = \pm\frac{\pi}{4}+k\pi,\ k=0,1,2,\cdots\ $ with lines parallel to the horizontal plane at a height of $z = 1$. The point $z=0$ behaves as a maximum for a curve's class and as a minimum for another curves so this point is a saddle point. The maxima and minima are located at the bordering circle or at $r = 1$ and those points are located at
$$ \cases{ \text{maxima}\to 2\theta = 2k\pi\\ \text{minima}\to 2\theta=(2k+1)\pi } $$
On
A geometrical interpretation of this problem can also be helpful. Since we are looking for level curves for the function, $ \ f(x,y) \ = \ e^{x^2 \ - \ y^2} \ = \ c \ \ , $ we could work with the curve equation $ \ x^2 \ - \ y^2 \ = \ \ln c \ \ . $ This is the (square) "horizontal" hyperbola centered on the origin, with its major axis on the $ \ x-$ axis and with asymptotes $ \ y\ = \ \pm x \ $ .
Its vertices then lie at $ \ ( \ \pm \sqrt{\ln c} \ , \ 0 ) \ \ . $ The largest possible value of $ \ \ln c \ $ for which the hyperbola will make contact with the boundary of the disk, $ \ x^2 \ + \ y^2 \ = \ 1 \ , $ is thus $ \ \ln c \ = \ 1 \ \ , $ so the absolute maximum value of our function occurs for $ \ f(\pm 1 \ , \ 0) \ = \ e \ \ . $
As we reduce $ \ \ln c \ $ , the vertices close in on the origin, reducing the hyperbola to its "degenerate form" of the two asymptotes. The two vertices have merged with the center of the hyperbola at $ \ (0 \ , \ 0 ) \ $ for $ \ \ln c \ = \ 0 \ \ \Rightarrow \ \ c \ = \ 1 \ . $
If we then continue to reduce $ \ \ln c \ $ by making it negative, the curve equation becomes
$$ \ x^2 \ - \ y^2 \ = \ \ln c \ \ \ \rightarrow \ \ \ y^2 \ - \ x^2 \ = \ -| \ \ln c \ | \ \ , $$
which is a (square) "vertical" hyperbola with the same center and asymptotes, but now with its major axis on the $ \ y-$ axis. The hyperbola will be just tangent to the boundary for $ \ -| \ \ln c \ | \ = \ 1 \ \ \Rightarrow \ \ \ln c \ = \ -1 \ \ . $ Hence, the absolute minimum value of the function occurs for the vertices $ \ f(0 \ , \ \pm 1) \ = \ e^{-1} \ \ . $
In the neighborhood of the origin, as $ \ \ln c \ $ increases from values slightly less than zero to those slightly greater than zero, the two points on the $ \ y-$ axis draw closer together, merge at the origin for $ \ c \ = \ 1 \ $ , and then separate along the $ \ x-$ axis. Thus, the critical point at the origin behaves as a "saddle point".
First of all, you solve the unconstrained optimization problem on the interior of the disk and find that the origin is the only critical point. Of course, it turns out to be a saddle point.
Now consider the Lagrange multiplier problem with the constraint $g(x,y) = x^2+y^2 = 1$. Then $\nabla f = \lambda\nabla g$ if and only if $$2xe^{x^2-y^2} = \lambda (2x) \quad\text{and}\quad -2ye^{x^2-y^2} = \lambda(2y).$$ Assuming $x,y\ne 0$, we deduce that $\lambda=e^{x^2-y^2} = -\lambda$. As you pointed out, this is impossible. Therefore, we are left with the critical points where $x=0$ and $y=0$ on the unit circle (where one of the equations drops out and there is a solution of the other). We have $f(\pm 1,0) = e$ and $f(0,\pm 1) = e^{-1}$, so the absolute maximum points are $(\pm 1,0)$ and the absolute minimum points are $(0,\pm 1)$.