Sketch the graph of $ f (x_1, \: x_2) = x_2 $ over $ x_1 ^ 2 + x_2 ^ 2 \leq 2 $ and find the maximum value of $ f $.

Question

Sketch the graph of $ f (x_1, \: x_2) = x_2 $ over $ x_1 ^ 2 + x_2 ^ 2 \leq 2 $ and find the maximum value of $ f $.

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The graph would be what's inside the cylinder, I think. My affirmation is that the maximum is $x_2 = \sqrt{2}$.

Answer 1

As said in comments, $f(x_1,x_2)=x_2$ is a linear function in two variables, so its graph is a plane. More precisely, the domain $D$ of the function here is the disk $D=\{\; (x_1,x_2)\; |\; x_1 ^ 2 + x_2 ^ 2 \leq 2\; \}$, so the graph is only the part of the plane $x_3=x_2$ above that disk.

My affirmation is that the maximum is $\sqrt{2}$.

Yes, but it should be proved. Clearly, since $(x_1,x_2)\in D$, we have $f(x_1,x_2)=x_2\le \sqrt{2}$. Otherwise, if $x_2>\sqrt{2}$, then $x_1^2+x_2^2 \ge x_2^2 > 2$. Since, $f(0,\sqrt{2})=\sqrt{2}$ and $(0,\sqrt{2})\in D$, $\sqrt{2}$ is indeed the maximum value of $f$ on $D$.