$$\begin{array}{ll} \text{maximize} & f(x,y) := x^{2} – y^{2}\\ \text{subject to} & g(x,y) := 1 – x^{2} – y^{2} = 0\end{array}$$
I tried solving this question using Lagrange multiplier and I had set up the following Lagrangean,
$$ L = x^{2} – y^{2} – \lambda (1 – x^{2} – y^{2}) $$
I got the following first-order conditions:
$$\frac{\partial L}{\partial x} = x + \lambda x \\ \frac{\partial L}{\partial y} = \lambda y – y $$
I am not sure how to solve after this.
When I equate the FOC to 0 I get x= 0 and y=0 but if I keep x= 1 and y = 0 a larger value for f(x,y) is obtained which still satisifes the constraint.
Best Answer
As your constraint is unit circle $x^2+y^2=1$, it is easy to see that you have maxima when the objective function is a hyperbola $x^2 - y^2 = 1$ tangent to the circle and you have minima when it is hyperbola $x^2-y^2 = -1$, again tangent to the circle.
As far as Lagrange Multiplier method, you have
$x(\lambda+1) = 0$
$y(\lambda - 1) = 0$
$1 - x^2 - y^2 = 0$
From first equation, $\lambda = -1$ is a possible solution. Plugging into second, $y = 0$ and from third, $x = \pm1$.
Also from second, $\lambda = 1$ is a possible solution. That gives you $x = 0$ and $y = \pm1$.
That leads to maxima of $f(x, y) = x^2-y^2$ as $1$ and minima as $-1$.
You also have $x = 0$ from the first equation and $y = 0$ from the second but they both lead to the same set of solutions.