# To find: The alternate solution to the exercise with the help of Lagrange Multiplier. x+2y+3z=6 Question
Alternate coordinate systems To find: The alternate solution to the exercise with the help of Lagrange Multiplier. $$\displaystyle{x}+{2}{y}+{3}{z}={6}$$ 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 $$\displaystyle{\frac{{{4}}}{{{3}}}}$$

### Relevant Questions Consider the following linear transformation T : P_2 \rightarrow P_3, given by T(f) = 3x^2 f'. That is, take the first derivative and then multiply by 3x^2 (a) Find the matrix for T with respect to the standard bases of P_n: that is, find [T]_{\epsilon}^{\epsilon}, where- \epsilon = {1, x, x^2 , x^n) (b) Find N(T) and R(T). You can either work with polynomials or with their coordinate vectors with respect to the standard basis. Write the answers as spans of polynomials. (c) Find the the matrix for T with respect to the alternate bases: [T]_A^B where A = {x - 1, x, x^2 + 1}, B = {x^3, x, x^2, 1}. 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. All bases considered in these are assumed to be ordered bases. In Exercise, compute the coordinate vector of v with respect to the giving basis S for V. $$\displaystyle{V}{i}{s}{P}_{{1}},{S}={\left\lbrace{t}+{1},{t}-{2}\right\rbrace},{v}={t}+{4}$$ All bases considered in these are assumed to be ordered bases. In Exercise, compute the coordinate vector of v with respect to the giving basis S for V. V is $$\displaystyle{M}_{{22}},{S}=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}\backslash{0}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{0}\backslash{1}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{1}\backslash{0}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{0}\backslash{0}&{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},{v}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}\backslash-{1}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}.$$ All bases considered in these are assumed to be ordered bases. In Exercise, compute the coordinate vector of v with respect to the giving basis S for V. V is $$\displaystyle{R}^{{2}},{S}=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}\backslash{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}\backslash{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},{v}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{3}\backslash-{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\$$ To determine:
a) The origin $$\displaystyle{\left[\begin{array}{cc} {0}&\ {0}\end{array}\right]}$$ is a critical point of the systems.
$$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}{\quad\text{and}\quad}{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$. Futhermore, it is a center of the corresponding linear system.
b) The systems $$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}{\quad\text{and}\quad}{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$ are almost linear.
c) To prove: $$\displaystyle{\left[{\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}\ {<}\ {0}\right]}{\quad\text{and}\quad}{\left[{r}\rightarrow\ {0}\ {a}{s}\ {t}\rightarrow\ \infty\right]},$$ hence the critical point for the system $$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ -\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ -\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$ is asymptotically stable and the solution of the initial value problem for $$\displaystyle{\left[{r}\ {w}{i}{t}{h}\ {r}={r}_{{{0}}}\ {a}{t}\ {t}={0}\right]}$$ becomes unbounded as $$\displaystyle{\left[{t}\rightarrow{\frac{{{1}}}{{{2}}}}\ {r}{\frac{{{2}}}{{{0}}}}\right]}$$, hence the critical point for the system $$\displaystyle{\left[{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={y}\ +\ {x}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)},\ {\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}=\ -{x}\ +\ {y}{\left({x}^{{{2}}}\ +\ {y}^{{{2}}}\right)}\right]}$$ is unstable. Given the elow bases for R^2 and the point at the specified coordinate in the standard basis as below, (40 points) $$\displaystyle{B}{1}=\le{f}{t}{\left\lbrace{\left({1},{0}\right)},{\left({0},{1}\right)}{r}{i}{g}{h}{t}\right\rbrace}&{M}{S}{K}{B}{2}={\left({1},{2}\right)},{\left({2},-{1}\right)}{r}{i}{g}{h}{t}\rbrace{\left({1},{7}\right)}={3}^{\cdot}{\left({1},{2}\right)}-{\left({2},{1}\right)}{N}{S}{K}{B}{2}={\left({1},{1}\right)},{\left(-{1},{1}\right)}{\left({3},{7}={5}^{\cdot}{\left({1},{1}\right)}+{2}^{\cdot}{\left(-{1},{1}\right)}{N}{S}{K}{B}{2}={\left({1},{2}\right)},{\left({2},{1}\right)}{\left({0},{3}\right)}={2}^{\cdot}{\left({1},{2}\right)}-{2}^{\cdot}{\left({2},{1}\right)}{N}{S}{K}{\left({8},{10}\right)}={4}^{\cdot}{\left({1},{2}\right)}+{2}^{\cdot}{\left({2},{1}\right)}{N}{S}{K}{B}{2}={\left({1},{2}\right)},{\left(-{2},{1}\right)}{\left({0},{5}\right)}={N}{S}{K}{\left({1},{7}\right)}=\right.}$$ a. Use graph technique to find the coordinate in the second basis. (10 points) b. Show that each basis is orthogonal. (5 points) c. Determine if each basis is normal. (5 points) d. Find the transition matrix from the standard basis to the alternate basis. (15 points) All bases considered in these are assumed to be ordered bases. In Exercise, compute coordinate vector v with respect to the giving basis S for V. V is $$\displaystyle{M}_{{22}},{S}=\le{f}{t}{\left\lbrace{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&-{1}\backslash{0}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{1}\backslash{1}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}\backslash{0}&-{1}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace},{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{0}\backslash-{1}&{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{r}{i}{g}{h}{t}\right\rbrace},{v}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{1}&{3}\backslash-{2}&{2}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}.$$ To calculate: The intercepts on the coordinate axes of the straight line with the given equation $$\displaystyle{2}{y}-{4}={3}{x}$$ 