Maximize $x^2+y^2+z^2$ on $x^2+y^2+4z^2 = 1$

Question

Hi this is a lagrangian optimization problem. Essentially as the title says, the question is asking us to maximize (if possible) $x^2+y^2+z^2$ on $x^2+y^2+4z^2=1$.

I started by the standard lagrangian method:$$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(1-x^2-y^2-4z^2)$$ Which subsequently gives: $${∂L\over∂x}=2x-2x\lambda=0$$ $${∂L\over∂y}=2y-2y\lambda=0$$ $${∂L\over∂y}=2z-8z\lambda=0$$

So the equations all reduce to: $\lambda=1$, $\lambda=1$, $\lambda=\frac14$ respectively.
I don't quite know how to proceed from here since generally it seems as though the above equations are supposed to reduce to something with x, y, or z so they can be plugged back into the original function.

Thanks a lot for the help, I'm sure the solution is somewhat trivial and I'm overlooking something simple.

Answer 1

If you must use Lagrange Multipliers, notice if $\lambda=1$ then the third equation forces $z=0$, hence

$$x^2+y^2+4(0)^2=1 \iff x^2+y^2=1 $$

Hence $f(x,y,z) = x^2+y^2+0^2 = 1$ is the maximum with no additional restriction on $x,y$.

If $\lambda=\dfrac{1}{4}$, then the first two equations force $x=y=0$, and

$$0^2+0^2+4z^2=1 \iff z=\pm\dfrac{1}{2} $$

This gives $f(x,y,z) = \dfrac{1}{4}$, which is less than the other one we found.

Hence $1$ is the maximum, and occurs along the unit circle in the plane $z=0$.

Answer 2

Hint: Rearrange the equation to get $x^2 + y^2 + z^2 = 1 - 3z^2$. Thus, we can instead maximize $1 - 3z^2$ on $x^2 + y^2 + 4z^2 = 1$.

Answer 3

$x^2+y^2+z^2=(1-4z^2)+z^2=1-3z^2$, whose value is maximal for $z=0$, in which case x and y describe the unit circle, centered in the origin.