Maximum value of $f(x, y, z) = e^{xyz}$ in the domain x+y+z = 3.

Question

Need the maximum value of f$(x, y, z) = e^{xyz}$ in the domain $x+y+z = 3$.

Please help me on this, as I could not find how to solve this one.

Thanks in advance.

Answer 1

Assuming $x,y,z \ge 0$

$e^{xyz}$ is maximum if $xyz$ is maximum

$AM \ge GM\\ \implies \dfrac{x+y+z}{3} \ge\sqrt[3]{xyz}\\ \implies \sqrt[3]{xyz} \le 1\\ \implies xyz \le 1$

Maximum value of $e^{xyz}=e^{1}=e$

Answer 2

For e^xyz the domain is all (x,y,z) ,so the domain is whole space