Find a function f(x, y) such that f is continuous everywhere.  the contour map of f consists of ellipses centered at the origin.  (0,0, f(0, 0)) is the highest point on the surface z = f(x, y).  Then find the range of f(x, y). Justify your answer using algebraic work.

The two variable function which is continuous everywhere with contours being ellipses centered at the origin and $(0, 0, f(0,0\left)\right)$ being the highest point will be upper ellipsoid.

The general equation of the ellipsoid is $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$, where $a,b,c$ are positive non-zero real numbers.

Thus, to get the function we can solve the equation for $z$ as follows:

$$\begin{gathered} \frac{z^2}{c^2}=1-\frac{x^2}{a^2}-\frac{y^2}{b^2} \\ \Rightarrow \frac{z}{c}=\pm \sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} \\ \Rightarrow z=\pm c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} \end{gathered}$$

so, we get two functions $f(x,y)=c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} \text{and} f(x,y)=-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}$.

The function with highest point $\left(0, 0, f(0,0)\right)$ is $f(x,y)=c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}$.