Maximize $f(x,y) = x^{2} - y^{2}$ subject to $g(x,y) = 1 - x^{2} - y^{2} = 0$

Question

$$\begin{array}{ll} \text{maximize} & f(x,y) := x^{2} - y^{2}\\ \text{subject to} & g(x,y) := 1 - x^{2} - y^{2} = 0\end{array}$$

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.

Answer 1

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.

Answer 2

Elementary option

Constraint: $y^2=1-x^2;$

$f(x,y)=x^2-y^2=2x^2-1;$

1)Min: at $x=0,$ and $y=\pm1:$

$f_{min}(x,y)=-1;$

2)Max: at $x=\pm1$ and $y=0:$

$f_{max}(x,y)=1.$