 # 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. Jaden Easton 2021-01-28 Answered
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,
$yz=\frac{xz}{2}$