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I'm having difficulty finding the minimum of this equation, $f(x,y,z) = xy + 2xz + 3yz$, subject to the constraint $xyz = 6$ and $x \ge 0$, $y \ge 0$, $z \ge 0$. I tried using Lagrange multipliers but got stuck when eliminating $\lambda$, and finding the values of $x, y, z$. Anyone has any idea how to solve this? Would appreciate any help!

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    $\begingroup$ You will have $ \ \lambda \ = \ \frac{y \ + \ 2z}{yz} \ = \ \frac{x \ + \ 3z}{xz} \ = \ \frac{2x \ + \ 3y}{xy} \ $ $ \Rightarrow \ \frac{1}{z} \ + \ \frac{2}{y} \ \ = \ \ \frac{1}{z} \ + \ \frac{3}{x} \ \ = \ \ \frac{2}{y} \ + \ \frac{3}{x} \ $ . You can work with each of the three implied equations to find proportions between $ \ x \ $ , $ \ y \ $ and $ \ z \ $ , which you then use in the constraint equation. (Incidentally, zero is not in the domain of any of these variables.) $\endgroup$ – colormegone Jan 14 '15 at 3:04
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    $\begingroup$ Ah, thanks so much! I get it now! $\endgroup$ – inggumnator Jan 14 '15 at 3:16
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Here is the basic definition of lagrange multipliers: $$ \nabla f = \lambda \nabla g$$ With respect to: $$ g(x,y,z)=xyz-6=0$$ Which turns into: $$\nabla (xy+2xz+3yz) = <y+2z,x+3z,2x+3y>$$ $$\nabla (xyz-6) = <yz,xz,xy>$$ Therefore, separating into components gives the following equations: $$ \vec i:y+2z=\lambda yz \rightarrow \lambda = \frac{y+2z}{yz}$$ $$ \vec j:x+3z=\lambda xz \rightarrow \lambda = \frac{x+3z}{xz}$$ $$ \vec k:2x+3y=\lambda xy \rightarrow \lambda = \frac{2x+3y}{xy}$$ Notice how I solved for lambda first thing. You now have a 3-way equality, which can be separated into two equations, each equaling zero: $$\frac{y+2z}{yz}-\frac{x+3z}{xz} = 0$$ $$\frac{y+2z}{yz}-\frac{2x+3y}{xy} = 0$$ Put each over a common denominator to simplify. $$xy+2xz-(xy+3yz)=2xz-3yz=z(2x-3y)=0$$ $$xy+2xz-(2xz+3yz)=xy-3yz=y(x-3z)=0$$ Notice how I can factor out a variable in each equation above. This gives us quite a collection of solutions: $$y=0,z=0$$ Plug into constraint: $$x(0)(0)-6=0 \rightarrow x \in \mathbb{R}$$

For the others, solve the equations: $$2x=3y \rightarrow x=\frac{3y}{2}$$ $$x-3z=0 \rightarrow x=3z$$ $$y=2z$$ Plug these into the constraint equation $g$, and you can find the location: $$(3z)(2z)z-6=6z^3-6=0 \rightarrow z=1$$ Therefore, the point of interest is: $$x=3,y=2,z=1$$ Which is exactly the answer that @PeterWoolfitt obtained in a much shorter way. This, however, uses Lagrange Multipliers.

As a general note, my strategy is to eliminate the $\lambda$ first, just because it is the new variable, and multiplied onto every equation in only one place. This can make the rest of the equations harder though, since you end up with lots of ratios.

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  • $\begingroup$ I think you can safely discard zero as the value for any of the variables, since the constraint equation gives a non-zero product for $ \ xyz \ $ . (The constraint surface approaches any of the coordinate planes asymptotically.) $\endgroup$ – colormegone Jan 14 '15 at 3:35
  • $\begingroup$ Good catch, that's kind of what I was coming to, since the constraint would allow infinitely many solutions if any one coordinate was $0$. $\endgroup$ – FundThmCalculus Jan 14 '15 at 5:34
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Warning: This method is cute, but does not use Lagrange multipliers.

Let $x=3w$ and $y=2u$. Then we have $xyz=6\implies uwz=1$ and we want to minimize $$xy+2xz+3yz=6uw+6wz+6uz=6(uw+wz+uz)$$ Now by the AM-GM inequality, we have $$uw+wz+uz\ge3\sqrt[3]{uwz}=3,$$ with equality when $uw=wz=uz$ which implies $u=w=z=1$, so we have $$xy+2xz+3yz\ge6\times3=18$$ and we can see $18$ is a true minimum because it is achieved at $x=3$, $y=2$, and $z=1$.

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  • $\begingroup$ This is a very clever solution. Doesn't answer the question in the exact method asked, but certainly cure and worth posting. Thanks! It also provided me a way to check my solution. $\endgroup$ – FundThmCalculus Jan 14 '15 at 3:18
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    $\begingroup$ This approach "re-scales" the variables in a way that creates a symmetry among the three variables $ \ w \ , \ u \ , \ $ and $ \ z \ $ . The condition of equality in the improper inequality expression points out the line of symmetry of the geometrical arrangement in $ \ wuz-$ space, $ \ w \ = \ u \ = \ z \ $ . This is a useful principle for solving a variety of "extremization" problems. (While it doesn't use "Lagrange", this technique can get you to the solution faster...) $\endgroup$ – colormegone Jan 14 '15 at 3:44

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