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I am attempting to find the absolute extremes of the function: $$f(x,y,z) = xyz$$ with the condition that: $$x+y+z=1$$

So far I have gathered the following:

Condition: $$C(x,y,z) = x+y+z-1$$ and the main function: $$F(x,y,z,\lambda) = xyz-\lambda x - \lambda y - \lambda z + \lambda$$

then calculating the derivatives: $$\frac{\partial F}{\partial x} = yz-\lambda \\ \frac{\partial F}{\partial y} = xz-\lambda \\ \frac{\partial F}{\partial z} = xy-\lambda \\ \frac{\partial F}{\partial \lambda} = -x-y-z=1$$

From here, how do I proceed?

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    $\begingroup$ are the variables assumed to be positive? $\endgroup$ – Dr. Sonnhard Graubner Nov 21 '17 at 15:53
  • $\begingroup$ @Dr.SonnhardGraubner the instructions says nothing about that. $\endgroup$ – Omari Celestine Nov 21 '17 at 15:54
  • $\begingroup$ Consider $f(x,x,1-2x) = x^2 (1-2x)$. This can take any value in $\mathbb{R}$. $\endgroup$ – copper.hat Nov 21 '17 at 15:56
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Let $y=x\rightarrow+\infty$.

Thus, $f\rightarrow-\infty.$

Let $x=y\rightarrow-\infty.$

Thus, $f\rightarrow+\infty.$

Id est, our function has no absolute extremum.

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