Finding maximum possible value

Question

Let $ x $, $ y $ and $ z $ be non-negative real numbers, $ y \ne 0 $, and let $ 2 x + \frac 3 y + \frac z 6 = 24 $. Determine the maximum possible value of $$ \sqrt [ 3 ] { \frac { x z } y } \text . $$

In class, we recently learned about inequalities between the arithmetic mean, geometric mean, harmonic mean, and quadratic mean. I started by stating that the given equation is less than or equal to $ \frac { x + z + \frac 1 y } 3 $. However, I don't know where to go from here. How can I relate the two equations we are given to finish this problem?

Answer 1

Direct resolution:

We have $$\frac{18}y=144-12x-z$$

and we can equivalently maximize

$$xz\left(144-12x-z\right).$$

By canceling the gradient,

$$\begin{cases}144z-24xz-z^2=0,\\144x-12x^2-2xz=0.\end{cases}$$

Both equations can be factored and the largest solution is obtained with the nonzero solutions.

Answer 2

Hint:by AM-GM$$8=\frac{2x+\frac{3}{y}+\frac{z}{6}}{3}\ge \sqrt[3]{2x.\frac{3}{y}.\frac{z}{6}}=\sqrt[3]{\frac{xz}{y}}$$