Let $D=\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2 \leq z \leq 9\}$ and $f:D > \rightarrow \mathbb{R}, f(x,y,z) = x+2y+\alpha z$.
For each value of $\alpha \in \mathbb{R}$, find maximum of $f$.
I started finding critical points at $D^{\mathrm{o}}$, taking the points $x \in D$ such that $\nabla f(x) = (0,0,0)$. As the gradient is $(1,2,\alpha)$, any point satisfy this so i got no critical points from here.
Now i want to see on the borders, so first i look at $z =9$. Considering $h(x,y) = f(x,y,9) = x+2y+9 \alpha$ we have that $\nabla h(x,y) \neq (0,0)$ for every $(x,y) $ st $ x^2+y^2 = z$. So there i got none critical points from here either.
Then i see around the border of the paraboloid, $x^2+y^2 = z$. If $g(x,y,z) = x^2+y^2-z$, then i could consider the set $C=\{(x,y,z): g(x,y,z) = 0\}$. As all the hypothesis for Lagrange Multipliers hold ($f,g $ are continuously differentiable and $\nabla g(x,y,z) = (2x, 2y, 1) \neq (0,0,0)$ for each $(x,y,z) \in D$) then $x \in D$ is extremum of $f$ restricted to $C$ if $\nabla f(p) = \lambda \nabla g(p)$. Solving the system yields the solution $(\frac{1}{-2\alpha}, \frac{1}{-\alpha}, \frac{5}{4\alpha^2})$ ($\alpha \neq 0$).
Now i have to add all the points in the intersection of $x^2+y^2 = z$ and $z=9$ and determine the maximum of the function comparing with these values, but i got stuck here. Any suggestions?