Maximum value of a function on the intersection between the surfaces

Question

I'm trying to determine the maximum value of $$ f(x,y,z) = x+y+z$$ On the intersection between $$ x^2 + y^2 +z^2 = 2 \quad \text{and} \quad x^2+y^2-z=0 $$ I tried to set the 2 functions equal and got $$ z^2+z-2=0 $$ But if i try to use Lagrange function, i.e $F(x,y,z,\lambda) = f(x,y,z)+\lambda (z^2+z-2)$ i won't get any information, how should i think here?

Answer 1

The problem can be here simplified. Firstly, we can substitute $z$ in $f$ and in the sphere constraint, that is, we want now to maximize

$$x+y+x^2+y^2$$ subject to $x^2+y^2+(x^2+z^2)^2=2$. Now, we can observe that the problem is symmetric and that the cost is an increasing function of $x,y$, which means that the optimal solution is attained for $x=y$. This means that we need new to maximize

$$2x+2x^2$$ subject to $2x^2+4x^4=2$.

Now, we can easily find the solution $x=\sqrt{2}/2$, which yields the maximum $1+\sqrt{2}$ with $x=y=\sqrt{2}/2$ and $z=1$.