Mazimisation problem with Lagrange's mutipliers

Question

$ Let f(x,y,z) = 4x^{1/4}y^{1/4}z^{1/4} $

Given that f is concave , maximize f(x,y,z) over R^{3} subject to $ x+y = 3$ and $ y+z = 3 $

$L(x,y,z , \lambda) = 4x^{1/4}y^{1/4}z^{1/4} + \lambda_{1}(3-x-y) + \lambda_{2}(3-y-z) $

end up with 5 equations in 5 unknowns, however I am so lost as to how to solve this, $ \Delta_{x}L=$

$ x^{-3/4}y^{1/4}z^{1/4} - \lambda_{1} = 0$

,$ x^{1/4}y^{-3/4}z^{1/4} - \lambda_{1} - \lambda_{2} = 0 $

$ x^{-1/4}y^{1/4}z^{-3/4} - \lambda_{2} = 0 $

$ x +y = 3$

$y+z=3$

from the last two equations I get x = z , so I can eliminate z, however this only gives me

$ x^{-1/2}y^{1/4} = \lambda_{1}$

$x^{1/2}y^{-3/4} -\lambda_{1}-\lambda_{2} =0 $

$ x^{-1/2}y^{1/4} = \lambda_{2} $

I just cant solve this from here, keep going round in circles trying to eliminate variables

Answer 1

$ x^{-1/2}y^{1/4} = \lambda_{1}$

$x^{1/2}y^{-3/4} -\lambda_{1}-\lambda_{2} =0 $

$ x^{-1/2}y^{1/4} = \lambda_{2} $

Plug the first and last into the middle one: $$x^{1/2}y^{-3/4} -x^{-1/2}y^{1/4}-x^{-1/2}y^{1/4} =0$$ For $xy \ne 0$, multiply by $x^{1/2}y^{3/4}$ to get: $$x-y-y =0 \iff x = 2y$$ Now check the different cases and remember that $x=z$:

• $x=0=z$, then by $x+y=3 \implies y = \ldots$
• $y=0$, then by $x+y=3 \implies x = \ldots = z$
• $x=2y$, then...

Answer 2

substitution from equations 1 and 3 in 2 concludes that $x=z=2$ and $y=1$

Answer 3

By AM-GM $$4\sqrt[4]{xyz}=4\sqrt[4]{\frac{1}{2}x\cdot2y\cdot z}\leq4\sqrt[4]{\frac{1}{2}\left(\frac{x+2y+z}{3}\right)^3}=4\sqrt2.$$ The equality occurs for $x+y=y+z=3$ and $x=2y=z$, id est, for $(x,y,z)=(2,1,2)$, which gives that $4\sqrt2$ is a maximal value.