I'm looking at Stewart's "Calculus" (8th ed). See here:
What confuses me is that the constraint $g(x,y) = x^2 + y^2 = 1$ is a cylinder, not a 2-d circle. Figure 2 above seems to be combining the graph of $z=f(x,y)$ with the graph of the level curve of $g(x,y)=1$.
Is Stewart just saving space here or am I missing something?
Stewart wants you to evaluate the function on the circle itself, i.e. in $\mathbb{R}^2$. The picture is in 3D to give you intuition, I guess, even though formally you would only need the values of $f$ on that circle, not anywhere inside...
Incidentally, that gives you a way to compute the extrema using 1D calculus. Parameterize the circle as $(x,y) = (\cos t, \sin t)$ for $0 \le t < 2\pi$, and now you seek the extrema of $$ f(x(t), y(t)) = x(t)^2 + 2y(t)^2 = \cos^2t + 2\sin^2t = 1 + \sin^2t $$ over $t \in [0,2\pi)$...