Max & Min Values of $S$ for $x_1^2 + x_2^2 = y_1^2 + y_2^2 = 2013$

[anemone]

Find the maximum and the minimum values of $S = (1 - x_1)(1 -y_1) + (1 - x_2)(1 - y_2)$ for real numbers $x_1, x_2, y_1,y_2$ with $x_1^2 + x_2^2 = y_1^2 + y_2^2 = 2013$.

 

[jvicens]

Hi there,

To find the maximum and minimum values of $S$, we can use the method of Lagrange multipliers. This method involves finding the critical points of a function subject to some constraints.

In this case, our function is $S = (1 - x_1)(1 -y_1) + (1 - x_2)(1 - y_2)$ and our constraints are $x_1^2 + x_2^2 = y_1^2 + y_2^2 = 2013$.

Using Lagrange multipliers, we can set up the following equations:

$\frac{\partial S}{\partial x_1} = \lambda \frac{\partial (x_1^2 + x_2^2)}{\partial x_1}$
$\frac{\partial S}{\partial x_2} = \lambda \frac{\partial (x_1^2 + x_2^2)}{\partial x_2}$
$\frac{\partial S}{\partial y_1} = \lambda \frac{\partial (y_1^2 + y_2^2)}{\partial y_1}$
$\frac{\partial S}{\partial y_2} = \lambda \frac{\partial (y_1^2 + y_2^2)}{\partial y_2}$
$x_1^2 + x_2^2 = y_1^2 + y_2^2 = 2013$

Solving these equations, we get the following critical points:

$(x_1, x_2, y_1, y_2) = (\pm \sqrt{2013}, 0, 0, \pm \sqrt{2013})$ and $(x_1, x_2, y_1, y_2) = (0, \pm \sqrt{2013}, \pm \sqrt{2013}, 0)$.

Plugging these values into our function $S$, we get the maximum value of $S = 4$ when $(x_1, x_2, y_1, y_2) = (\pm \sqrt{2013}, 0, 0, \pm \sqrt{2013})$ and the minimum value of $S = 0$ when $(x_1, x_2, y_1, y_2) = (