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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.

PM.
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Sivabalanarayanan L
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    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. – QuantumPotatoïd Jan 15 '17 at 19:24

2 Answers2

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

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

Tayyab
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  • The question already specified the domain and asked about the maximum value of the function. Typesetting the math in your answer would also help. – PM. Apr 04 '17 at 15:41