# The volume of the largest rectangular box which lies in the first octant with three faces in the coordinate planes and its one of the vertex in the plane displaystyle{x}+{2}{y}+{3}{z}={6} by using Lagrange multipliers.

The volume of the largest rectangular box which lies in the first octant with three faces in the coordinate planes and its one of the vertex in the plane $x+2y+3z=6$ by using Lagrange multipliers.
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Definition used:
" The Lagrange multipliers defined as $\mathrm{\nabla }f\left(x,y,z\right)=\lambda \mathrm{\nabla }g\left(x,y,z\right).$ This equation can be expressed as ${f}_{x}=\lambda {g}_{x},{f}_{y}=\lambda {g}_{y},{f}_{z}=\lambda {g}_{z}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(x,y,z\right)=k$"
Calculation:
Let the volume of the rectangular box be
Thus, the maximize function $f\left(x,y,z\right)=xyz$ subject to the constraint $g\left(x,y,z\right)=x+2y+3z=6.$
The Lagrange multipliers $\mathrm{\nabla }f\left(x,y,z\right)=\lambda \mathrm{\nabla }g\left(x,y,z\right)$ is computed as follows,
$\mathrm{\nabla }f\left(x,y,z\right)=\lambda \mathrm{\nabla }g\left(x,y,z\right)$
$⟨{f}_{x},{f}_{y},{f}_{z}⟩=\lambda ⟨{g}_{x},{g}_{y},{g}_{z}⟩$
$⟨\begin{array}{c}{f}_{x}\left(xyz\right)\\ {f}_{y}\left(xyz\right)\\ {f}_{z}\left(xyz\right)\end{array}⟩=\lambda ⟨\begin{array}{c}{g}_{x}\left(x+2y+3z\right)\\ {g}_{y}\left(x+2y+3z\right)\\ {g}_{z}\left(x+2y+3z\right)\end{array}⟩$
$⟨yz,xz,xy⟩=\lambda ⟨1,2,3⟩$
Thus, the value of $\mathrm{\nabla }f\left(x,y,z\right)=\lambda \mathrm{\nabla }g\left(x,y,z\right)is⟨yz,xz,xy⟩=\lambda ⟨1,2,3⟩.$
By the definition, $⟨yz,xz,xy⟩=\lambda ⟨1,2,3⟩$ can be expressed as follows,
$yz=\lambda \left(1\right)$
$xz=2\lambda \left(2\right)$
$xy=3\lambda \left(3\right)$
From the equations (1), (2) and (3),
$yz=\frac{xz}{2}=\frac{xy}{3}$
Consider $yz=\frac{xz}{2}$ and compute the value of y,
$yz=\frac{xz}{2}$