To find: The alternate solution to the exercise with the help of Lagrange Multiplier. x+2y+3z=6

Question
Alternate coordinate systems
asked 2021-02-19
To find: The alternate solution to the exercise with the help of Lagrange Multiplier. \(\displaystyle{x}+{2}{y}+{3}{z}={6}\)

Answers (1)

2021-02-20
Method used: In order to find the maximum and minimum values of a function f(x, y, z) that is subjected to the constraint\(\displaystyle{g{{\left({x},{y},{z}\right)}}}={k}\) such that \(\displaystyle\triangle{d}{o}{w}{n}{g}\ne{q}{0}.\) a) Find the values x, y, z and \(\displaystyle\lambda\) when \(\displaystyle\triangle{d}{o}{w}{n}{f{{\left({x},{y},{z}\right)}}}=\lambda\triangle{d}{o}{w}{n}{g{{\left({x},{y},{z}\right)}}}\) and \(\displaystyle{g{{\left({x},{y},{z}\right)}}}={k}\) b) Compute the values at the points from step a and find the maxomum and minimum values of f. Calculation: It is given that the plane is \(\displaystyle{x}+{2}{y}+{3}{z}={6}\) that lies in the first octant. Consider a point (x,y,z) on the plane. Then, the function becomes PSKf(x,y,z)=xyz. Here, \(\displaystyle{g{{\left({x},{y}\right)}}}={x}+{2}{y}+{3}{z}\). Fint \(\displaystyle{f}_{{{x}}}\ \text{as}\ {{f}_{{{x}}}{\left({x},{y},{z}\right)}}={y}{z},{f}_{{{y}}}\ \text{as}\ {{f}_{{{y}}}{\left({x},{y},{z}\right)}}={x}{z},{f}_{{{z}}}\ \text{as}\ {{f}_{{{z}}}{\left({x},{y},{z}\right)}}={x}{y},{g}_{{{x}}}\ \text{as}\ {{g}_{{{x}}}{\left({x},{y},{z}\right)}}={1},{g}_{{{y}}}\ \text{as}\ {{g}_{{{y}}}{\left({x},{y},{z}\right)}}={2}\text{and}{g}_{{{z}}}\text{as}{{g}_{{{z}}}{\left({x},{y},{z}\right)}}={3}\) Thus, the equations are obtained as follows. \(\displaystyle{{f}_{{{x}}}{\left({x},{y},{z}\right)}}=\lambda{{g}_{{{x}}}{\left({x},{y},{z}\right)}}\)
\(\displaystyle{y}{z}=\lambda{\left({1}\right)}\)
\(\displaystyle{y}{z}=\lambda\) Solve the equation \(\displaystyle{{f}_{{{y}}}{\left({x},{y},{z}\right)}}={x}{z}\) as shown below. \(\displaystyle{{f}_{{{y}}}{\left({x},{y},{z}\right)}}=\lambda{{g}_{{{y}}}{\left({x},{y},{z}\right)}}\)\
\(\displaystyle{x}{z}=\lambda{\left({2}\right)}\)
\(\displaystyle{x}{z}={2}\lambda\)
\(\displaystyle{\frac{{{x}{z}}}{{{2}}}}=\lambda\) Solve the equation \(\displaystyle{{f}_{{{z}}}{\left({x},{y},{z}\right)}}={x}{y}\) as follows. \(\displaystyle{{f}_{{{z}}}{\left({x},{y},{z}\right)}}=\lambda{{g}_{{{z}}}{\left({x},{y},{z}\right)}}\)
\(\displaystyle{x}{y}=\lambda{\left({3}\right)}\)
\(\displaystyle{x}{y}={3}\lambda\)
\(\displaystyle{\frac{{{x}{y}}}{{{3}}}}=\lambda\) Then, \(\displaystyle{y}{z}={\frac{{{x}{z}}}{{{2}}}}={\frac{{{x}{y}}}{{{3}}}}=\lambda\) Consider two terms at a time, take \(\displaystyle{y}{z}={\frac{{{x}{z}}}{{{2}}}}\) that gives \(\displaystyle{y}{z}={\frac{{{x}{z}}}{{{2}}}}\)
\(\displaystyle{y}={\frac{{{x}}}{{{2}}}}\) Then take \(\displaystyle{y}{z}={\frac{{{x}{y}}}{{{3}}}}\) that gives \(\displaystyle{y}{z}={\frac{{{x}{y}}}{{{3}}}}\)
\(\displaystyle{z}={\frac{{{x}}}{{{3}}}}\) Substitute the values of y and z in the equation \(\displaystyle{x}+{2}{y}+{3}{z}={6}\) and obtain the value of x as follows. \(\displaystyle{x}+{2}{\left({\frac{{{x}}}{{{2}}}}\right)}+{3}{\left({\frac{{{x}}}{{{3}}}}\right)}={6}\)
\(\displaystyle{x}+{x}+{x}={6}\)
\(\displaystyle{3}{x}={6}\)
\(\displaystyle{x}={2}\) Then, the values of y and z are \(\displaystyle{y}={\frac{{{2}}}{{{2}}}}={1}\) and \(\displaystyle{z}={\frac{{{2}}}{{{3}}}}\). Thus, the point is \(\displaystyle{\left({2},{1},{\frac{{{2}}}{{{3}}}}\right)}\) Then, the function value at this point \(\displaystyle{\left({2},{1},{\frac{{{2}}}{{{3}}}}\right)}\) becomes f(2,1,\frac{2}{3})=(2)(1)(\frac{2}{3})=\frac{4}{3}. Therefore, the volume of the largest rectangular box whose one vertex lie the plane \(\displaystyle{x}+{2}{y}+{3}{z}={6}\) is \(\displaystyle{\frac{{{4}}}{{{3}}}}\)
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