Determine the maximum value of $f(x,y,z) = (xyz)^{(1/3)}$ given that x, y, and z are non-negative numbers that satisfy $x+y+z =1$

Question

Is Lagrange multipliers in this question appropriate? I don't really know how to begin solving this question..

Answer 1

Since AM $\geq$ GM,

$\frac{x+y+z}{3} \geq (xyz)^{\frac{1}{3}}$

Therefore

$(xyz)^{\frac{1}{3}} \leq \frac{1}{3}$

Alternatively you can do the following:

Let $f(x, y, z) = (xyz)^{\frac{1}{3}} + \lambda (x+y+z -1)$

Set $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial z} = 0$

Simplifying,

$x = y = z = \frac{1}{3}$

Hence the maximum value of $(xyz)^{\frac{1}{3}}$ is $\left( \left(\frac{1}{3} \right)^3 \right)^{\frac{1}{3}} = \frac{1}{3}$

One difficulty in this case is checking the second order condition. You'll have to compute the Bordered Hessian matrix. More specifically, in this case, you will find the details here, Theorem $5$, page $6$. If you compute everything properly, you'll find that $-\text{det}(H_4) = 3 > 0$ and $-\text{det}(H_3) = -2 < 0$ and hence this is a local maximum.