I want to find the extreme values of the function $f(x,y,z) = 2x + 2y + z$ under the constraints $x^2+y^2+z^2 \le 2$ and $x^2 + y^2 \le z$
The task is to use a parametrization of the two constraints/conditions to find stationary points of f on the boundary.
I don't even know where to begin with this - I can solve it using Lagrange but that is not what we're supposed to do on this particular assignment. Thankful for any help!
From this 3D graph you can see that the boundary of the constrained region has two parts: the bottom of the paraboloid $x^2+y^2=z$ for $0\le z\le 1$, and the cap of the sphere $x^2+y^2+z^2=2$ for $1\le z\le \sqrt 2$.
How did I get those limits for $z$? Equate the right-hand sides of the equations $x^2+y^2=2-z^2$ and $x^2+y^2=2$ to get $2-z^2=2$ which has $z=1$ as the only positive solution. The other limits $0$ and $\sqrt 2$ more obviously come from each equation.
We then parameterize those surfaces. For the bottom of the paraboloid,
$$x=\sqrt u\cos v$$ $$y=\sqrt u\sin v$$ $$z=u$$ $$\text{for}\quad 0\le u\le 1,\ 0\le v\le 2\pi$$
For the sphere's cap,
$$x=\sqrt{2-u^2}\cos v$$ $$y=\sqrt{2-u^2}\sin v$$ $$z=u$$ $$\text{for}\quad 1\le u\le \sqrt 2,\ 0\le v\le 2\pi$$
You should also check for optima on "the boundary of the boundary," the circle where the two parameterizations overlap. You can do that by taking $u=1$ in either parameterization. You get
$$x=\cos v$$ $$y=\sin v$$ $$z=1$$ $$\text{for}\quad 0\le v\le 2\pi$$
Since the region is convex and your function to optimize is linear, any optima will definitely be on the region's boundary and not in its interior.
You should be able to finish from here.