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?

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.