\hfill \thepage} %} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{Ma 116 Lecture 3/22/00} \vspace{1pt} \subsection{Lagrange Multipliers} Suppose we want to find the stationary values of a function $f\left( x,y\right) $, that is, the points at which a function $f\left( x,y\right) $ might have either a maximum or minimum, in the case when the two variables $% x $ and $y$ are not mutually independent, but are connected by a constraint of the form $\varphi \left( x,y\right) =0.$ \paragraph{\protect\vspace{1pt}} Suppose one wants to find the rectangle with given area $16$ that has the smallest perimeter. Then if $x$ and $y$ are the dimensions of the rectangle, we want to maximize $f\left( x,y\right) =2\left( x+y\right) $ subject to the fact that $xy=16.$ That is we want to maximize $f=2\left( x+y\right) $ subject to the condition $\varphi \left( x,y\right) =xy-16=0.$ \vspace{1pt} Suppose that the $\varphi =0$ curve and the level curves $f\left( x,y\right) =k$ curves in the $x,y-$plane look as below. \begin{center} \FRAME{dtbpF}{4.2947in}{2.0012in}{0pt}{}{}{lagrange1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 4.2947in;height 2.0012in;depth 0pt;original-width 3.333in;original-height 1.542in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'graphics/lagrange1.bmp';file-properties "XNPEU";}} \vspace{1pt} \end{center} As we describe the $\varphi =0$ we encounter curves $f=k,$ and in general $k$ changes monotonically, i.e., either increases or decreases. At the point where the sense in which we run through the $c-$scale is reverses we may expect an extremum value. From the figure this will occur at a point $\left( x_{0},y_{0}\right) $ where the $f=k$ \ curve and the $\varphi =0$ have the same tangent. Thus at $\left( x_{0},y_{0}\right) $ \vspace{1pt} \begin{equation*} f_{x}:f_{y}=\varphi _{x}:\varphi _{y}\text{ \ or }\dfrac{f_{x}}{\varphi _{x}}% =\dfrac{f_{y}}{\varphi _{y}}=-\lambda \end{equation*} \vspace{1pt} where $\lambda $ is a constant of proportionality. (Recall that the normal to $f\left( x,y\right) -k=0$ is $\nabla f=f_{x}\vec{i}+f_{y}\vec{j}$ so that in the $x,y-$plane a tangent vector is $-f_{y}\vec{i}+f_{x}\vec{j}.$ Thus we have the two conditions \vspace{1pt} \begin{eqnarray*} f_{x}+\lambda \varphi _{x} &=&0 \\ f_{y}+\lambda \varphi _{y} &=&0 \end{eqnarray*} \vspace{1pt} at the extremum point $\left( x_{0},y_{0}\right) .$ We also have the third condition $\varphi \left( x,y\right) =0.$ These three conditions allow us to solve for the three unknowns $x_{0},y_{0},$ and $\lambda .$ The constant $% \lambda $ is called a \emph{Lagrange multiplier.} \vspace{1pt} We may state Lagrange's Rule: To find the extreme values of the function $f\left( x,y\right) $ subject to the subsidiary condition $\varphi \left( x,y\right) =0,$ add to $f$ the product of $\varphi $ and an unknown factor $\lambda $ independent of $x$ and $y,$ and write down the known necessary conditions $f_{x}+\lambda \varphi _{x}=0$ and $f_{y}+\lambda \varphi _{y}=0$ for an extreme value of $% F=f+\lambda \varphi .$ These two equations in connection with the subsidiary condition $\varphi =0$ determine the coordinates of the extreme value and $% \lambda .$ \vspace{1pt} \begin{example} (Same as above.) Suppose one wants to find the rectangle with given area $16$ that has the smallest perimeter. \end{example} Then if $x$ and $y$ are the dimensions of the rectangle, we want to maximize $f\left( x,y\right) =2\left( x+y\right) $ subject to the fact that $xy=16.$ That is we want to maximize $f=2\left( x+y\right) $ subject to the condition $\varphi \left( x,y\right) =xy-16=0.$ \vspace{1pt} We form $F=2\left( x+y\right) +\lambda \left( xy-16\right) .$ Then \vspace{1pt} \begin{eqnarray*} F_{x} &=&2+\lambda y=0 \\ F_{y} &=&2+\lambda x=0 \end{eqnarray*} \vspace{1pt} Thus for $\lambda \neq 0$ we have that $x=y.$ Then $\varphi =0$ implies that $x=y=4,$ \ which is a square. \vspace{1pt} The method of Lagrange Multipliers may be applied to functions of three or more variables. \begin{example} \vspace{1pt}Find the points on \begin{equation*} 4x-5y+3z=2 \end{equation*} \end{example} which are closest to $(1,-2,3)$ \vspace{1pt} \emph{Solution:}\ \ \ The distance is given by: \begin{equation*} \sqrt{(x-1)^{2}+(y+2)^{2}+(z-3)^{2}} \end{equation*} It is easier to work with the square of the distance. Thus we want to minimize \begin{equation*} f(x,y,z)=(x-1)^{2}+(y+2)^{2}+(z-3)^{2} \end{equation*} subject to the constraint \begin{equation*} \varphi (x,y,z)=4x-5y+3z-2=0 \end{equation*} Let \begin{eqnarray*} F\left( x,y,z\right) &=&f\left( x,y,z\right) +\lambda \varphi \left( x,y,z\right) \\ &=&(x-1)^{2}+(y+2)^{2}+(z-3)^{2}+\lambda \left( 4x-5y+3z-2\right) \end{eqnarray*} \vspace{1pt} Then \ \ \ \begin{eqnarray*} F_{x} &=&2(x-1)+4\lambda =0 \\ F_{y} &=&2(y+2)-5\lambda =0 \\ F_{z} &=&2(z-3)+3\lambda =0 \end{eqnarray*} \ \ or% \begin{eqnarray*} \dfrac{x-1}{-2} &=&\lambda \\ \dfrac{2\left( y+2\right) }{5} &=&\lambda \\ \dfrac{2\left( z-3\right) }{-3} &=&\lambda \end{eqnarray*} Thus \begin{equation*} \dfrac{x-1}{-2}=\dfrac{2\left( z-3\right) }{-3} \end{equation*}% , Solution is: $\left\{ x=-3+\frac{4}{3}z\right\} $ and \begin{equation*} \dfrac{2\left( y+2\right) }{5}=\dfrac{2\left( z-3\right) }{-3} \end{equation*}% , Solution is: $\left\{ y=3-\frac{5}{3}z\right\} $ \vspace{1pt} But the point $\left( x,y,z\right) $ lies on the plane and must satisfy $% \varphi (x,y,z)=4x-5y+3z-2=0,$ so% \begin{equation*} 4\left( -3+\frac{4}{3}z\right) -5\left( 3-\frac{5}{3}z\right) +3z=2 \end{equation*}% , Solution is: $\left\{ z=\frac{87}{50}\right\} $ Thus \begin{equation*} x=-3+\frac{4}{3}\left( \dfrac{87}{50}\right) =-\frac{17}{25} \end{equation*} \begin{equation*} y=3-\frac{5}{3}\left( \dfrac{87}{50}\right) =\frac{1}{10} \end{equation*} \begin{example} Find the extreme values of $f\left( x,y\right) =xy$ on the circle $% x^{2}+y^{2}=1.$ \end{example} $\varphi \left( x.y\right) =x^{2}+y^{2}-1$ so that $F\left( x,y\right) =xy+\lambda \left( x^{2}+y^{2}-1\right) .$ thus we have the equations \vspace{1pt} \begin{eqnarray*} y+2\lambda x &=&0 \\ x+2\lambda y &=&0 \\ x^{2}+y^{2} &=&1 \end{eqnarray*}% , Solution is: $\left\{ \lambda =\frac{1}{2},y=\rho _{1},x=-\rho _{1}\right\} ,\allowbreak \left\{ y=\rho _{1},x=\rho _{1},\lambda =-\frac{1}{% 2}\right\} \allowbreak $ where $\rho _{1}$ is a root of $2\hat{Z}^{2}-1.$ Thus $x_{0}=y_{0}=\pm \dfrac{\sqrt{2}}{2}.$ We have the following results \vspace{1pt} \begin{center} \begin{tabular}{llll} $x_{0}$ & $y_{0}$ & $f\left( x_{0}y_{0}\right) $ & \\ & & & \\ $\dfrac{\sqrt{2}}{2}$ & $\dfrac{\sqrt{2}}{2}$ & $\dfrac{1}{2}$ & maximum \\ $\dfrac{\sqrt{2}}{2}$ & $-\dfrac{\sqrt{2}}{2}$ & $-\dfrac{1}{2}$ & minimum \\ $-\dfrac{\sqrt{2}}{2}$ & $\dfrac{\sqrt{2}}{2}$ & $-\dfrac{1}{2}$ & minimum \\ $-\dfrac{\sqrt{2}}{2}$ & $-\dfrac{\sqrt{2}}{2}$ & $\dfrac{1}{2}$ & maximum% \end{tabular} \vspace{1pt} \end{center} \begin{example} Suppose one wants to cut a beam with maximal rectangular cross section from a circular log of radius $\sqrt{2}.$ \end{example} We shall use Lagrange multipliers to show that the optimal beam has square cross section. Let the origin be at the center of the log and the beam so that the $x-$axis cuts the log and beam in half horizontally and the $y-$% axis cuts the log and beam in half vertically. (This means that the beam has dimensions $2x$ by $2y.)$ The log satisfies the equation% \begin{equation*} x^{2}+y^{2}=2 \end{equation*} \vspace{1pt} If $\left( x,y\right) $ is the corner of the beam in first quadrant, then we must maximize the area \begin{equation*} A=f\left( x,y\right) =4xy \end{equation*} of the beam's rectangular cross section subject to the constraint \begin{equation*} \varphi \left( x,y\right) =x^{2}+y^{2}-2=0 \end{equation*} Then% \begin{equation*} F\left( x,y\right) =4xy+\lambda \left( x^{2}+y^{2}-2\right) \end{equation*} \vspace{1pt} so that \begin{eqnarray*} F_{x} &=&4y+2\lambda x=0 \\ F_{y} &=&4x+2\lambda y=0 \end{eqnarray*} Hence% \begin{equation*} -\lambda =\dfrac{2y}{x}=\dfrac{2x}{y} \end{equation*} or% \begin{equation*} x^{2}=y^{2} \end{equation*} But the fact that the corner of the beam must lie on the log, that is the circle $x^{2}+y^{2}=2$ tells us that \begin{equation*} x^{2}+x^{2}=2 \end{equation*} so $x=y=1.$ (Recall that $\left( x,y\right) $ is in the first quadrant.) Thus the beam is square with dimensions $2x=2y=2.$ \vspace{1pt} \end{document} %%%%%%%%%%%%%%%%%%%% End /document/lec_3_22_00.tex %%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%% Start /document/graphics/lagrange1.bmp %%%%%%%%%%%%%%% BudOyB@@@@@@@xC@@@@J@@@@@E@@@PI@@@P@@D@@@@@@@@rE@@@@@@@@@@@@@H@@@@`@@@@@@@@ @@|Cp }w O|C ~_mCw {u~_oW{p} _}w} Cp_ } w O w o{ ~ }_ w } { ~ { ~ w~ w_x ~c~ g{ __o{ s_ O{|o _o o}}o_~ {|}w O{ |~oG }_~}w G{x w{{_oy op}_} so_|{g~p o{_G~{C _}owxp~C~_{ pGpo@|qg_ {_} |}CN@@@@@@@@@|xw {qo O}~_~G ~{s^s _yoOn_|c_U~ w@C|cw]{_~CC `q_wogwO@@@@@xDw~ g}ww{O oosw ~gw}{ |wo}o{_O|}_ ~oc}g{ sq}oo_~_x~ wg~gC~x|sA {ocAppo{ O@@@@@~g {{_o s~x {q oOO}~w G{{_a o{o_|c_~ O@C|{sC C`qoOO@@@@@ ~q{x oq ~O~{ s_~o}O~ ~yx{ ~O|os a~p~|O~O|`w {sO|A|{O@_|o _|~O~ {_|o c~g {c oO~w C__x oO||cCOx}k C_@~CpGw~ %%%%%%%%%%%%%%%%% End /document/graphics/lagrange1.bmp %%%%%%%%%%%%%%%%