I need help to find the maximum ($x^∗, y^∗, z^∗$) of the production function
$Q(x, y, z) = x^{1/4}y^{1/4}z^{1/4}$
subject to the budget constraint $h(x, y, z) = ax+by +cz −d = 0$, (where $a, b, c, d$ are positive constants), in terms of these constants. And from this, I must find an expression for the maximum value $Q^∗$ of the budget in terms of $a, b, c, d$ and the corresponding value $λ^∗$ of the Lagrange multiplier.
So far in this question I have set up a system of linear equations from the partial derivatives of the Lagrangian form:
$L_x = \dfrac{1}{4}y^{1/4}z^{1/4}y^{-3/4} – a\lambda = 0$
$L_y = \dfrac{1}{4}x^{1/4}z^{1/4}x^{-3/4} – b\lambda = 0$
$L_z = \dfrac{1}{4}x^{1/4}y^{1/4}z^{-3/4} – c\lambda = 0$
$-L_\lambda = ax + by +cz -d = 0$
From here I have trouble solving these equations and taking the question to the next step.
Best Answer
The Lagrangian relaxation can be written as $$ \text{Maximize } L(\lambda) = (xyz)^{1/4} + \lambda(d - ax - by - cz) \\ \lambda \geq 0 $$ Partially differentiating with respect to each variable, we get \begin{align} \frac{\partial L}{\partial x} = \frac{(yz)^{1/4}}{4x^{3/4}}-a\lambda = 0 \tag 1\\ \frac{\partial L}{\partial y} = \frac{(xz)^{1/4}}{4y^{3/4}}-b\lambda = 0 \tag 2\\ \frac{\partial L}{\partial z} = \frac{(xy)^{1/4}}{4z^{3/4}}-c\lambda = 0 \tag 3\\ \frac{\partial L}{\partial \lambda} = ax + by + cz - d = 0 \tag 4 \end{align}
Solving equations $(1)$ and $(2)$ we get $ax = by$ and solving equations $(2)$ and $(3)$ we get $by = cz$. Once we know this, we can find out the optimal values are
$$ x^* = \frac{d}{3a} \\ y^* = \frac{d}{3b} \\ z^* = \frac{d}{3c} \\ \lambda^* = \Big(\frac{3}{abcd}\Big)^{1/4} \\ $$