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 displaystyle{x}+{2}{y}+{3}{z}={6} by using Lagrange multipliers.

Question
Alternate coordinate systems
asked 2021-01-28
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.

Answers (1)

2021-01-29
Definition used:
" The Lagrange multipliers defined as \(\displaystyle\nabla f{{\left({x},{y},{z}\right)}}=\lambda\nabla g{{\left({x},{y},{z}\right)}}.\) This equation can be expressed as \(\displaystyle{{f}_{{x}}=}\lambda{{g}_{{x}},}{{f}_{{y}}=}\lambda{{g}_{{y}},}{{f}_{{z}}=}\lambda{{g}_{{z}}{\quad\text{and}\quad}} g{{\left({x},{y},{z}\right)}}={k}\)"
Calculation:
Let the volume of the rectangular box be \(\displaystyle{V}= f{{\left({x},{y},{z}\right)}}={x}{y}{z}\ \text{where}\ {x}>{0},{y}>{0},{z}>{0}.\)
Thus, the maximize function \(\displaystyle f{{\left({x},{y},{z}\right)}}={x}{y}{z}\) subject to the constraint \(\displaystyle g{{\left({x},{y},{z}\right)}}={x}+{2}{y}+{3}{z}={6}.\)
The Lagrange multipliers \(\displaystyle\nabla f{{\left({x},{y},{z}\right)}}=\lambda\nabla g{{\left({x},{y},{z}\right)}}\) is computed as follows,
\(\displaystyle\nabla f{{\left({x},{y},{z}\right)}}=\lambda\nabla g{{\left({x},{y},{z}\right)}}\)
\(\displaystyle{\left\langle{{f}_{{x}},}{{f}_{{y}},}{f}_{{z}}\right\rangle}=\lambda{\left\langle{{g}_{{x}},}{{g}_{{y}},}{g}_{{z}}\right\rangle}\)
\(\displaystyle{\left\langle\begin{matrix}{{f}_{{x}}{\left({x}{y}{z}\right)}}\\{{f}_{{y}}{\left({x}{y}{z}\right)}}\\{{f}_{{z}}{\left({x}{y}{z}\right)}}\end{matrix}\right\rangle}=\lambda{\left\langle\begin{matrix}{{g}_{{x}}{\left({x}+{2}{y}+{3}{z}\right)}}\\{{g}_{{y}}{\left({x}+{2}{y}+{3}{z}\right)}}\\{{g}_{{z}}{\left({x}+{2}{y}+{3}{z}\right)}}\end{matrix}\right\rangle}\)
\(\displaystyle{\left\langle{y}{z},{x}{z},{x}{y}\right\rangle}=\lambda{\left\langle{1},{2},{3}\right\rangle}\)
Thus, the value of \(\displaystyle\nabla f{{\left({x},{y},{z}\right)}}=\lambda\nabla g{{\left({x},{y},{z}\right)}}{i}{s}{\left\langle{y}{z},{x}{z},{x}{y}\right\rangle}=\lambda{\left\langle{1},{2},{3}\right\rangle}.\)
By the definition, \(\displaystyle{\left\langle{y}{z},{x}{z},{x}{y}\right\rangle}=\lambda{\left\langle{1},{2},{3}\right\rangle}\) can be expressed as follows,
\(\displaystyle{y}{z}=\lambda{\left({1}\right)}\)
\(\displaystyle{x}{z}={2}\lambda{\left({2}\right)}\)
\(\displaystyle{x}{y}={3}\lambda{\left({3}\right)}\)
From the equations (1), (2) and (3),
\(\displaystyle{y}{z}=\frac{{{x}{z}}}{{2}}=\frac{{{x}{y}}}{{3}}\)
Consider \(\displaystyle{y}{z}=\frac{{{x}{z}}}{{2}}\) and compute the value of y,
\(\displaystyle{y}{z}=\frac{{{x}{z}}}{{2}}\)
\(\displaystyle{y}=\frac{1}{{2}}{x}\)
Considder \(\displaystyle\frac{{{x}{z}}}{{2}}=\frac{{{x}{y}}}{{3}}\) and compute the value of z,
\(\displaystyle\frac{{{x}{z}}}{{2}}=\frac{{{x}{y}}}{{3}}\)
\(\displaystyle{z}=\frac{2}{{3}}{y}\)
Substitute \(\displaystyle{y}=\frac{1}{{2}}{x}{\quad\text{and}\quad}{z}=\frac{2}{{3}}{y}\) in the equation \(\displaystyle g{{\left({x},{y},{z}\right)}}={x}+{2}{y}+{3}{z}={6}\) and obtain required point,
\(\displaystyle{x}+{2}{y}+{3}{z}={6}\)
\(\displaystyle{x}+{2}{\left(\frac{1}{{2}}{x}\right)}+{3}{\left(\frac{2}{{3}}\cdot\frac{1}{{2}}{x}\right)}={6}\)
\(\displaystyle{x}+{x}+{x}={6}\)
\(\displaystyle{x}={2}\)
Thus, the values of \(\displaystyle{x}={2},{y}={1}{\quad\text{and}\quad}{z}=\frac{2}{{3}}.\)
Substitute \(\displaystyle{x}={2},{y}={1}{\quad\text{and}\quad}{z}=\frac{2}{{3}}\) in the function \(\displaystyle f{{\left({x},{y},{z}\right)}}={x}{y}{z}\) and obtain the maximum volume of the rectangle box as follows,
\(\displaystyle f{{\left({x},{y},{z}\right)}}={x}{y}{z} f{{\left({2},{1},\frac{2}{{3}}\right)}}={\left({2}\right)}{\left({1}\right)}{\left(\frac{2}{{3}}\right)}\)
\(\displaystyle=\frac{4}{{3}}\)
Therefore, the volume of the largest rectangular box is \(\displaystyle\frac{4}{{3}}.\)
0

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