Concerning Lagrange undetermined multipliers

Question

$$\frac{3}{x} +\frac{4}{y} +\frac{5}{z} =6$$ Find minimum value of $x+y+z$?

I'm stuck with algebraic part of solving the equations. Please help me out.

Answer 1

To use Lagrange multipliers, set $$F(x,y,z)=x+y+z+\lambda \left({3\over x}+{4\over y}+{5\over z}\right)$$ Then $$ \begin{align} {\partial F\over\partial x} &= 1-{3\lambda\over x^2}\\ {\partial F\over\partial y} &= 1-{4\lambda\over y^2}\\ {\partial F\over\partial z} &= 1-{5\lambda\over z^2}\\ \end{align}$$ Setting the partial derivative equal to $0$ gives $$ \begin{align} \lambda &= 3x^2\\ \lambda &= 4y^2\\ \lambda &= 5z^2\\ \end{align}$$ so that $$ \begin{align} y &= \pm{\sqrt{3}x\over2}\\ z &= \pm{\sqrt{5}x\over\sqrt{3}}\\ \end{align}$$ This gives $4$ substitutions to make into $\frac{3}{x} +\frac{4}{y} +\frac{5}{z} =6$ to find local extrema.

Answer 2

The minimal value does not exist.

Try $x\rightarrow-\infty$, $y=4$ and $z\rightarrow1$.

For positive variables we obtain: $$x+y+z=\frac{1}{6}(x+y+z)\left(\frac{3}{x}+\frac{4}{y}+\frac{5}{z}\right)\geq\frac{1}{6}(\sqrt3+\sqrt4+\sqrt5)^2.$$ The equality occurs when $$(\sqrt{x},\sqrt{y},\sqrt{z})||\left(\sqrt{\frac{3}{x}},\frac{2}{\sqrt{y}},\sqrt{\frac{5}{z}}\right),$$ which says that we got a minimal value.

Indeed, let $$(\sqrt{x},\sqrt{y},\sqrt{z})=k\left(\sqrt{\frac{3}{x}},\frac{2}{\sqrt{y}},\sqrt{\frac{5}{z}}\right).$$ Thus, $$(x,y,z)=(k\sqrt3,2k,k\sqrt5),$$ which says $$\frac{\sqrt3+2+\sqrt5}{k}=6$$ or $$k=\frac{\sqrt3+2+\sqrt5}{6}$$ and we see that the equality occurs for $$(x,y,z)=\left(\frac{\sqrt3(\sqrt3+2+\sqrt5)}{6},\frac{2(\sqrt3+2+\sqrt5)}{6},\frac{\sqrt5(\sqrt3+2+\sqrt5)}{6}\right).$$