Find x, y, and z (all of them non-negative) that maximize $ x^\frac14 y^\frac14 z^\frac14 $ subject to the constraint that $ px + y + z ≤ m $ Find x, y, and z (all of them non-negative) that maximize $ x^\frac14 y^\frac14 z^\frac14 $ subject to the constraint that $ px + y + z ≤ m $
Context: undergraduate mathematical economics
I have attempted this problem to the best of my ability but I'm really not sure if I got the mechanics of it right. If somebody would be kind enough to check my work over and point out any potential errors I would be greatly appreciative.
Setting up the Lagrangian:
$ L = x^\frac14 y^\frac14 z^\frac14 - \lambda (px + y + z - m) ≤ 0 $
First Order Conditions:
$ f_x = \frac14x^\frac{-3}4 y^\frac14 z^\frac14 - \lambda p $
$ f_y = x^\frac14 \frac14y^\frac{-3}4 z^\frac14 - \lambda $
$ f_z = x^\frac14 y^\frac14 \frac14z^\frac{-3}4 - \lambda $
$ f_\lambda = -px-y-z+m $
So if I'm not mistaken we'd then get $ px = y = z $ by simplifying the first three equations.
Then, using $ f_\lambda = -px-y-z+m $, we can obtain expressions for x, y, and z:
$ x = \frac m{3p} $
$ y = \frac m3 $
$ z = \frac m3 $
Second part of the problem involves using the envelope theorem to solve for $ \frac {dV}{dm} $. Using the above I calculated this to be:
$ \frac {dV}{dm} = \frac{\partial L}{\partial m} = \lambda $
Taking $ f_z = x^\frac14 y^\frac14 \frac14z^\frac{-3}4 = \lambda $
=> $ \frac{m}{3p}^\frac14 \frac{m}{3}^\frac14 \frac14 \frac{m}{3}^\frac{-3}{4} = \lambda $
$ \lambda = \frac{3^\frac14}{4(m^\frac14)(p^\frac14)} $
Am I on the right track here? Been a while since I've taken a class involving this kind of math so I'm a little rusty.
 A: Another approach is to recognize that $(x y z)^{1/4}$ is maximized when $(p x y z)^{1/3}$ is maximized and apply the AM-GM inequality:
$$(p x y z)^{1/3} \le \frac{p x + y + z}{3} = \frac{m}{3},$$
with equality when $px=y=z$, yielding $$(x,y,z)=\left(\frac{m}{3p},\frac{m}{3},\frac{m}{3}\right).$$
A: If $p$ and $m$ are also nonnegative, we might as well assume $px+y+z=m$.  The Lagrange method now directs us to solve the following equations.
$$
\left\{\begin{array}{l}
\frac{\partial}{\partial x}(x^{1/4}y^{1/4}z^{1/4})=\lambda\frac{\partial}{\partial x}(px+y+z)
\\
\frac{\partial}{\partial y}(x^{1/4}y^{1/4}z^{1/4})=\lambda\frac{\partial}{\partial y}(px+y+z)
\\
\frac{\partial}{\partial z}(x^{1/4}y^{1/4}z^{1/4})=\lambda\frac{\partial}{\partial z}(px+y+z)
\\
px+y+z=m\end{array}\right.
$$
Simplifying gives
$$
\left\{\begin{array}{l}
\frac{1}{4}x^{-3/4}y^{1/4}z^{1/4}=\lambda p
\\
\frac{1}{4}x^{1/4}y^{-3/4}z^{1/4}=\lambda
\\
\frac{1}{4}x^{1/4}y^{1/4}z^{-3/4}=\lambda\\
px+y+z=m\end{array}\right.
$$
and hence
$$
\left\{\begin{array}{l}
x^{-3/4}y^{1/4}z^{1/4}=4\lambda p
\\
x^{1/4}y^{-3/4}z^{1/4}=4\lambda
\\
x^{1/4}y^{1/4}z^{-3/4}=4\lambda\\
px+y+z=m\end{array}\right.
$$
Now let's raise the first three equations to the fourth power, to obtain
$$
\left\{\begin{array}{l}
x^{-3}yz=256\lambda^4p^4
\\
xy^{-3}z=256\lambda^4
\\
xyz^{-3}=256\lambda^4\\
px+y+z=m\end{array}\right.
$$
The middle two equations give $xy^{-3}z=xyz^{-3}$ and hence $y=z$, which should be no surprise since both functions are symmetric in $y$ and $z$.  Hence
$$
\left\{\begin{array}{l}
x^{-3}y^2=256\lambda^4p^4
\\
xy^{-2}=256\lambda^4
\\
px+2y=m\end{array}\right.
$$
Making another obvious substitution gives
$$
\left\{\begin{array}{l}
x^{-3}y^2=xy^{-2}p^4
\\
px+2y=m\end{array}\right.
$$
You can take it from here I'm sure.
