Level set of functions $f(x,y) = x+y$ and $g(x,y) = xy$ with $M := \{ (x,y) \in \mathbb{R^2}: x^2+y^2 \leq 9 \}$

Question

Given are the functions

$$f(x,y) = x+y \\ g(x,y) = xy$$

and the set $$M := \{ (x,y) \in \mathbb{R^2}: x^2+y^2 \leq 9 \}$$

The question was to visualize the level set and to mark the points

$$\max_{(x,y)\in M} f(x,y) \\ \max_{(x,y)\in M} g(x,y)$$

I know that the set is a circle with the center in its origin and a radius of $3$, but I don't know how one can calculate the maximum here.

Apparently the red dots here mark the corresponding solution. Can someone please elaborate how we get to those?

Answer 1

The lines that are plotted in the background all have one thing in common: The objective function is constant on those lines (you see the value of the objective function by the color).

Moreover, if you were to plot every such "level line", you would get exactly R^2. So what you do is to find the line of the brightest color that intersects with the admissible set. In the first picture, of course, you can't see such line, but that does not mean it is not existent. It is simply not plotted. What you also notice is that the diagonal lines are getting brighter and brighter the more you increase $x$ and $y$. This is very obvious since your objective function increases as $x$ and $y$ increases. So just imagine where the brightest line would be and then find its intersection with the admissible set. This is where the objective function attains its highest value.

Maybe it would be helpful if you were to command Maxima to plot more lines, but I do not know this command by heart unfortunately.

Answer 2

In this context it is convenient to write $f$ and $g$ in terms of polar coordinates. Note that $M$ is given by points $(x,y) = (r\cos \alpha, r\sin \alpha)$ with $r \leq 3$. Also \begin{align*} f(x,y) & = r(\sin \alpha + \cos \alpha) = r\sqrt 2 \left(\frac{\sqrt 2}{2}\sin \alpha + \frac{\sqrt 2}{2}\cos \alpha\right) = \sqrt 2 r \sin\left(\frac{\pi}{4}+\alpha\right), \\ g(x,y) & = r^2\sin \alpha \cos \alpha = \frac12 r^2 \sin 2\alpha. \end{align*} Note that $\sin(\pi/4+\alpha) = 1$ when $\pi/4+\alpha = \pi/2$, so $\alpha = \pi/4$. The same is true for $\sin 2\alpha$. Hence $f$ and $g$ reach their maximum for $r=3$ and $\alpha = \pi/4$, i.e. at $(x,y) = (3\sqrt2/2, 3\sqrt 2/2)$. You can then use the symmetry of $f$ and $g$ to find other maximum points.