Finding minimum using Lagrange multipliers

Question

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!

Answer 1

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.

Answer 2

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 $z=1$.