Maximization problem

I can't solve it by Algebra but by Calculus using Lagrange multipliers and defining $F(x,y,z,\lambda) = xy+2yz - \lambda(x^2+y^2+z^2 - 36)$ . Then taking partial derivatives and equating them to zero.

$\begin{array} {ll} \dfrac {\partial F}{\partial x} = y - 2\lambda x & \implies \dfrac yx = 2 \lambda \\ \dfrac {\partial F}{\partial y} = x+2z - 2\lambda y & \implies \dfrac {x+2z}y = 2 \lambda \\ \dfrac {\partial F}{\partial z} = 2y - 2\lambda z & \implies \dfrac {2y}z = 2 \lambda \\ \dfrac {\partial F}{\partial \lambda} = - x^2 - y^2 - z^2 + 36 & \implies x^2+y^2+z^2 = 36 \end{array}$

$\implies 2 \lambda = \dfrac yx = \dfrac {x+2z}y = \dfrac {2y}z$ . From $\dfrac yx = \dfrac {2y}z \implies z = 2x$ .

From $\dfrac yx = \dfrac {x+2z}y = \dfrac {5x}y \implies y^2 = 5x^2$ . From $x^2+y^2+z^2 = 36 \implies x^2 + 5x^2 + 4x^2 = 36$

$\implies x = \dfrac 6{\sqrt {10}}$ , $y= 3\sqrt 2$ , and $z = \dfrac {12}{\sqrt{10}}$ .

Therefore, $\max (F(x,y,z,0) = F \left(\dfrac 6{\sqrt {10}}, 3\sqrt 2, \dfrac {12}{\sqrt{10}}, 0\right) = \dfrac 6{\sqrt {10}} \cdot 3\sqrt 2 + 2 \cdot 3\sqrt 2 \cdot \dfrac {12}{\sqrt{10}} = 18\sqrt 5 \approx \boxed{40.2}$