%ho147.tex: (2nd ed.) %%a Plain TeX file by Doron Zeilberger for Math 251(1 page) %\bigskip & \bigskip \cr %& %\cr %begin macros \def\L{{\cal L}} \baselineskip=14pt \parskip=10pt \def\Tilde{\char126\relax} \font\eightrm=cmr8 \font\sixrm=cmr6 \font\eighttt=cmtt8 \magnification=\magstephalf \def\P{{\cal P}} \def\Q{{\cal Q}} \parindent=0pt \overfullrule=0in \def\frac#1#2{{#1 \over #2}} %end macros \centerline { \bf Dr. Z's Math251 Handout \#14.7 (2nd ed.) [Optimization in Several Variables] } By Doron Zeilberger {\bf Problem Type 14.7a}: Find the local maximum and minimum values and saddle point(s) of the function $f(x,y)$. {\bf Example Problem 14.7a}: Find the local maximum and minimum values and saddle point(s) of the function $$ f(x,y)=x^3y+12x^2-8y \quad . $$ \bigskip \hrule \bigskip %\halign{#\hfill & \quad #\hfill \cr \halign{ \vtop{\hsize=15pc\noindent #\strut} \qquad & \vtop{\hsize=15pc\noindent #\strut} \cr {\bf Steps} & {\bf Example} \cr & \cr {\bf 1.} Find the partial derivatives $f_x$ and $f_y$. For future reference, also compute $f_{xx}$, $f_{xy}$ and $f_{yy}$. & {\bf 1.} $$ f_x={{\partial} \over {\partial x}}(x^3y+12x^2-8y)=3x^2y+24x \quad , $$ $$ f_y={{\partial} \over {\partial y}}(x^3y+12x^2-8y)=x^3-8 \quad , $$ $$ f_{xx}={{\partial} \over {\partial x}}(3x^2y+24x)=6xy+24 \quad , $$ $$ f_{xy}={{\partial} \over {\partial y}}(3x^2y+24x)=3x^2 \quad , $$ $$ f_{yy}={{\partial} \over {\partial y}}(x^3-8)=0 \quad . $$ \cr \bigskip & \bigskip \cr {\bf 2.} Set both $f_x$ and $f_y$ to zero. Then solve the system of two equations and two unknowns. The solutions are the {\bf critical points}. & {\bf 2.} $$ 3x^2y+24x=0 \quad, \quad x^3-8=0 \quad . $$ Which is the same as $$ x(3xy+24)=0 \quad, \quad x^3-8=0 \quad . $$ From the second equation we get $x=2$, and plugging it into the first we get \hfill\break $2 \cdot (6y+24)=0$, so $y=-4$. It turns out that in this problem there is only one critical point: $(2,-4)$. \cr \bigskip & \bigskip \cr {\bf 3.} Plug the point(s), one at a time, into $f_{xx},f_{xy}, f_{zz}$ and for each compute the {\bf discriminant} $D=f_{xx}f_{yy}-[f_{xy}]^2$. At the examined point $(a,b)$: If $D>0$ and $f_{xx}>0$ then $(a,b)$ is a {\bf local minimum} and the local minimum value is $f(a,b)$. If $D>0$ and $f_{xx}<0$ then $(a,b)$ is a {\bf local maximum} and the local maximum value is $f(a,b)$. If $D<0$, then $(a,b)$ is neither max. nor min but a {\bf saddle point}. If $D=0$ then we {\bf don't know} (the test is inconclusive). & {\bf 3.} $$ f_{xx}(2,-4)=6(2)(-4)+24=-24 \quad , $$ $$ f_{xy}(2,-4)=3\cdot 2^2=12 \quad , $$ $$ f_{yy}(2,-4)=0 \quad . $$ Since $D=(-24) \cdot 0-12^2=-144$ is {\bf negative}, this is a {\bf saddle point}. {\bf Ans.}: The function has no maximum values and no mimimum values. It has one saddle point at $(2,-4)$. \cr} \vfill\eject {\bf Problem Type 14.7b}: Find the absolute maximum and minimum values of $f$ on the set $D$ $$ f(x,y)=Expression(x,y) \quad , $$ $$ S=\{(x,y) \, \vert \, a_1 \leq x \leq a_2 \, , \, b_1 \leq y \leq b_2 \, \} \quad . $$ {\bf Example Problem 14.7b}: Find the absolute maximum and minimum values of $f$ on the set $S$ $$ f(x,y)=4x+6y-x^2-y^2 \quad , $$ $$ S=\{(x,y) \, \vert \, 0 \leq x \leq 4 \, , \, 0 \leq y \leq 5 \, \} \quad . $$ \bigskip \hrule \bigskip %\halign{#\hfill & \quad #\hfill \cr \halign{ \vtop{\hsize=15pc\noindent #\strut} \qquad & \vtop{\hsize=15pc\noindent #\strut} \cr {\bf Steps} & {\bf Example} \cr & \cr {\bf 1.} First find the {\bf critical points} by computing $f_x$, $f_y$, setting them both equal to $0$, and solving for $x$ and $y$. Only retain those points that belong to $S$. Then plug-in these point(s) into $f$, and keep them for comparison later on. %%%End General Step 1 & {\bf 1.} $f_x=4-2x$, $f_y=6-2y$. Solving $$ 4-2x=0 \quad , \quad 6-2y=0 \quad , $$ gives one solution $x=2,y=3$. So $(2,3)$ is a critial point. Also $f(2,3)=4 \cdot 2+6 \cdot 3-2^2-3^2=8+18-4-9=13$. \cr \bigskip & \bigskip \cr {\bf 2.} For each part of the boundary (in the case of a rectangle there are four sides), find the absolute max and min, like you did way back in Calc I. & {\bf 2.} On the LEFT side $x=0$ and $0 \leq y \leq 5$. $f(0,y)=6y-y^2$. Let's call this function, for now $F(y)$. $F'(y)=6-2y$ which is $0$ at $y=3$. $F(3)=6 \cdot 3 -3^2=18-9=9$. At the endpoints $F(0)=0, F(5)=5$. So looking at the numbers $0,3,9$, the largest is $9$ and the smallest is $0$ {\bf abs. min. on Left Side}: $0$, {\bf abs. max. on Left Side}: $9$. On the RIGHT side $x=4$ and $0 \leq y \leq 5$. $f(4,y)=16+6y-16-y^2=6y-y^2$. Let's call this function, for now $F(y)$. $F'(y)=6-2y$ which is $0$ at $y=3$. $F(3)=6 \cdot 3 -3^2=18-9=9$. At the endpoints $F(0)=0, F(5)=5$. So looking at the numbers $0,5,9$, the largest is $9$ and the smallest is $0$ {\bf abs. min. on Right Side}: $0$, {\bf abs. max. on Right Side}: $9$. On the DOWN side $y=0$ and $0 \leq x \leq 4$. $f(x,0)=4x-x^2$. Let's call this function, for now $F(x)$. $F'(x)=4-2x$ which is $0$ at $x=2$, and $F(2)=4$. At the endpoints $F(0)=0, F(4)=0$. So looking at the numbers $4,0,0$, the largest is $4$ and the smallest is $0$ {\bf abs. min. on DOWN Side}: $0$, {\bf abs. max. on DOWN Side}: $4$. On the UP side $y=5$ and $0 \leq x \leq 4$. $f(x,5)=4x-x^2+5$. Let's call this function, for now $F(x)$. $F'(x)=4-2x$ which is $0$ at $x=2$, and $F(2)=4$. At the endpoints $F(0)=5, F(4)=5$. So looking at the numbers $4,5,5$, the largest is $5$ and the smallest is $4$ {\bf abs. min. on UP Side}: $4$, {\bf abs. max. on UP Side}: $5$. \cr \bigskip & \bigskip \cr {\bf 3.} Now gather all these champions (in both min. and max. catergories) plus those came from the critical points inside the region and find the largest value, this is your {\bf absolute maximum value} and the smallest, this is your {\bf absolute minimum value}. & {\bf 3.} For abs. min the contdenders are $0,0,0,4,13$ so the {\bf absolute minimum value} is $0$. For abs. max the contdenders are $9,9,4,5,13$ so the {\bf absolute maximum value} is $13$. \cr} {\bf Problem Type 14.7c}: Find the point on the surface $F(x,y,z)=k$ that is closest to the origin. {\bf Example Problem 14.7c}: Find the point on the surface $x^2y^2z=1$ that is closest to the origin. \bigskip \hrule \bigskip %\halign{#\hfill & \quad #\hfill \cr \halign{ \vtop{\hsize=15pc\noindent #\strut} \qquad & \vtop{\hsize=15pc\noindent #\strut} \cr {\bf Steps} & {\bf Example} \cr & \cr {\bf 1.} It is more convenient to consider the distance-squared, which is $x^2+y^2+z^2$. Take one of the variables (say $z$) (whatever is convenient) and express it in terms of the other two (say $x,y$), and plug it into $x^2+y^2+z^2$ getting a function, let's call it $f(x,y)$ & {\bf 1.} $z=1/(x^2y^2)$ so $z^2=x^{-4}y^{-4}$ and the distance-squared, in terms of $x,y$ is $f(x,y)=x^2+y^2+x^{-4}y^{-4}$. \cr \bigskip & \bigskip \cr {\bf 2.} Find the critical points by taking $f_x,f_y$ and setting them equal to zero, and solving for $x$ and $y$. & {\bf 2.} $$ f_x= {{\partial} \over {\partial x}} (x^2+y^2+x^{-4}y^{-4})= 2x-4x^{-5}y^{-4} \quad , $$ $$ f_y= {{\partial} \over {\partial y}} (x^2+y^2+x^{-4}y^{-4})= 2y-4x^{-4}y^{-5} \quad . $$ We have to solve $$ 2x-4x^{-5}y^{-4}=0 \quad , \quad 2y-4x^{-4}y^{-5}=0 \quad , $$ which is the same $$ x^6y^4=2 \quad , \quad x^4y^6=2 \quad . $$ Dividing the first by the second we get $x^2/y^2=1$ so $x^2=y^2$ and $x^{10}=2$ and we get $x^2=y^2=2^{1/5}$ so $x=\pm 2^{1/10}$, $y=\pm 2^{1/10})$. \cr \bigskip & \bigskip \cr {\bf 3.} To get the $z$ coordinates for each of these points plug into $f(x,y)$. & {\bf 3.} $z=1/(x^2y^2)$ so for each of the four possibilities $z=1/2^{2/5}=2^{-2/5}$. {\bf Ans.:} The points on the surface closest to the origin are $(\pm 2^{1/10},\pm 2^{1/10},2^{-2/5})$. \cr} {\bf Problem from a Previous Final} Find the local maximum and minimum points, the local maximum and minimum values, and saddle points of the function $$ f(x,y)= 4x^2+y^2+2x^2y-1 \quad . $$ {\bf Ans.}: $(0,0)$ is the local minimum point $-1$ is the local minimum value $(\sqrt 2, -2), (-\sqrt 2, -2)$ are saddle points. {\bf Another Problem from a Previous Final} Find the local maximum and minimum point(s), the local maximum and minimum values, and saddle point(s) of the function $$ f(x,y)=6y^2-2y^3+3x^2+6xy \quad . $$ {\bf Ans.} Local max: none. Local Min: location: $(0,0)$, value: $0$. Saddle point: $(-1,1)$. \end