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

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

• Does youtr question mean $x,y, z > 0$? if no, maximum value is infinite – kotomord Jan 15 '17 at 18:53
• can you say something to the variables? – Dr. Sonnhard Graubner Jan 15 '17 at 18:56
• Since the exponential function is monotone increasing, maximizing $f(x,y,z)$ is equivalent to maximizing $xyz$, subject to the restriction $x+y+z = 3$. As kotomord points out, this cannot be maximized without further restriction on the variables. – hardmath Jan 15 '17 at 19:01
• I suppose it means find a local maximum with isolating one variable with x+y+z=3 then you get a function of two variables. It has a local maximum if the derivatives are all zero and the determinant of the hessian is negative. – Julien Pitteloud Jan 15 '17 at 19:24

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 $e^{xyz}=e^{1}=e$