User:Astronouth7303/LaGrange multipliers

This is a problem I'm working on for calc3 involving Lagrange multipliers.

Find the absolute minimum and maximum of $$f(x,y) = 16 - x^2 - y^2$$ given the constraint $$(x - 1)^2 + (y - 1)^2 = 4$$.

I've solved to here.

$$ \begin{cases} -2x = \lambda \cdot 2(x - 1) \\ -2y = \lambda \cdot 2(y - 1) \\ (x - 1)^2 + (y - 1)^2 = 4 \end{cases} $$

Ok, I think I got it:

$$-2x = \lambda \cdot 2(x - 1)$$

$$-x = \lambda (x - 1) \,$$

$$\frac{-x}{x-1} = \lambda$$

and similarly for y.

Put x and y together:

$$\frac{-x}{x-1} = \frac{-y}{y-1}$$

$$\frac{x}{x-1} = \frac{y}{y-1}$$

$$\frac{x-1}{x} = \frac{y-1}{y}$$

$$\frac{x}{x} - \frac{1}{x} = \frac{y}{y} - \frac{1}{y}$$

$$1 - \frac{1}{x} = 1 - \frac{1}{y}$$

$$\frac{-1}{x} = \frac{-1}{y}$$

$$\frac{1}{x} = \frac{1}{y}$$

$$y = x \,$$

Assuming:
 * $$x \ne 0$$
 * $$y \ne 0$$
 * $$x - 1 \ne 0$$
 * $$y - 1 \ne 0$$