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 \(\displaystyle\frac{4}{{3}}.\)
" 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 \(\displaystyle\frac{4}{{3}}.\)
