Extrema of implicit functions - one point - two values?!

Question

I am given this expression: $$x^2+y^2+z^2-2x-2y-2z+2 = 0$$ And I need to find the extrema of the function $$z = z(x,y)$$ I did it in the following way:

1. Differentiate the expression with respect to $x$ and solve for $\frac{dz}{dx}$
2. Differentiate with respect to $y$ and solve for $\frac{dz}{dy}$

3. Find all tuples $(x, y ,z)$ such that $\frac{dz}{dx} = \frac{dz}{dy} = 0$ given that both partial derivatives are continuous and exist.

4. Now, I got two points: $(1, 1, 0) $ and $(1, 1, 2)$ both of them turn out to be an extremum, but - to be honest - I do not know what is happening - why does this function have two outputs for one input?

Could you explain this to me in as simple terms as possible?

Answer 1

You have the equation of a sphere. When you calculate the extremum, you get the highest $z$ and the lowest one. You can rewrite the original equations as: $$(x-1)^2+(y-1)^2+(z-1)^2=1$$ You can write the equation for the upper hemisphere or the equation of the lower hemisphere

Answer 2

If solved for $(z-1)^2$ it is one function $f(x,y)$ of $x,y$ but to get just $z-1$ (and then just $z$) one has to use $\pm \sqrt{f(x,y)}.$ So really two functions.

Answer 3

You can solve your equation for $z$ $$z_1=1-\sqrt{-x^2+2 x-y^2+2 y-1}$$ and $$z_2=1+\sqrt{-x^2+2 x-y^2+2 y-1}$$ and differentiate this with respaect to $x,y$