Question

Найти наибольшее и наименьшее значение функции z=sin x + sin y + sin (x+y) в прямоугольнике 0 <= х <= π/2 ; 0 <= у <= π/2

Answer 1

Найдем частные производные:

$displaystyleLarge z=sin{x}+sin{y}+sin{(x+y)}\\{partial zoverpartial x}=cos{x}+cos{(x+y)},;{partial zoverpartial y}=cos{y}+cos{(x+y)}\begin{cases} &cos{x}+cos{(x+y)}=0\ &cos{y}+cos{(x+y)}=0 end{cases}\ cos{x}=cos{y}Rightarrow x=y\begin{cases} & cos{x}+cos{(2x)}=0 \ & cos{y}+cos{(2y)}=0 end{cases}\ cos{x}+cos^2{x}-sin^2{x}=0\ cos{x}+cos^2{x}-1+cos^2{x}=0\ 2cos^2{x}+cos{x}-1=0\ cos{x}=t,; tin[-1;1]\2t^2+t-1=0\D=1+8=9\t_1={-1+3over4}={1over2}\t_2={-1-3over4}=-1\ cos{x}={1over2}\ x_{1,2}=pm{piover3}+2pi n, ninmathbb{Z}\ cos{x}=-1\ x_{3}=pi+2pi k, ; kinmathbb{Z}\ cos{y}+cos^2{y}-sin^2{y}=0\ cos{y}+cos^2{y}-1+cos^2{y}=0\ 2cos^2{y}+cos{y}-1=0\ cos{y}=t,; tin[-1;1]\ 2t^2+t-1=0\ D=1+8=9\ t_1={-1+3over4}={1over2}\ t_2={-1-3over4}=-1\ cos{y}={1over2}\ y_{1,2}=pm{piover3}+2pi m, minmathbb{Z}\ cos{y}=-1\ y_{3}=pi+2pi c, ; cinmathbb{Z}$

Проверим принадлежность точек к нашей области:

$displaystyle D: begin{cases} & 0leq{x}leq{piover2}\ &0leq{y}leq{piover2} end{cases}\\ x_1={piover3}+2pi n,; ninmathbb{Z},; y_1={piover3}+2pi m,; minmathbb{Z}\ x_2=-{piover3}+2pi l,; linmathbb{Z},; y_2=-{piover3}+2pi w,; winmathbb{Z}\ x_3=pi+2pi k,; kinmathbb{Z},; y_3=pi+2pi c,; cinmathbb{Z} \ 0leq{piover3}+2pi nleq{piover2},; ninmathbb{Z}\ -{piover3}leq2pi nleq{piover2}-{piover3},; ninmathbb{Z}\ left(-{1over6}leq nleq{1over12},; ninmathbb{Z}right)Rightarrowmathbf{n=0}Rightarrow M_{0}left({piover3};{piover3}right)\ 0leq-{piover3}+2pi lleq{piover2},; linmathbb{Z}\ left({1over6}leq lleq{5over12},; linmathbb{Z}right)Rightarrowmathbf{lnotinmathbb{Z}}\ 0leqpi+2pi kleq{piover2},; kinmathbb{Z}\ left(-{1over2}leq kleq-{1over8},; kinmathbb{Z}right)Rightarrowmathbf{knotinmathbb{Z}}$

Найдем критические точки на границах(исходя из уравнений границ области):

$displaystyle mathbf{y_1=0}\ z=sin{x}+sin{x}=2sin{x}\ z'=2cos{x}\ 2cos{x}=0\ x_1={piover2}+pi n,; ninmathbb{Z}\\ mathbf{x_2=0}\ z=2sin{y}\ z'=2cos{y}\ 2cos{y}=0\ y_2={piover2}+pi k,; kinmathbb{Z}\\ mathbf{y_3={piover2}}\ z=sin{x}+cos{x}+1\ z'=cos{x}-sin{x}\ x_3={piover4}+pi m,; minmathbb{Z}\\ mathbf{x_4={piover2}}\ z=sin{y}+cos{y}+1\ z'=cos{y}-sin{y}\ y_4={piover4}+pi c,; cinmathbb{Z}\\ M_1left({piover2};0right),;;M_2left(0;{piover2}right),;;M_3left({piover4};{piover2}right),;;M_4left({piover2};{piover4}right)$

Также нужно проверить и граничные точки прямоугольника:

$displaystyle M_5left(0;0right),;;M_6left({piover2};{piover2}right)\\ z(M_0)={3sqrt{3}over2}\ z(M_1)=2 \ z(M_2)=2 \ z(M_3)=1+sqrt{2} \ z(M_4)=1+sqrt{2} \ z(M_5)=0 \ z(M_6)=2$

Сравним корни:

$displaystyle {3over2}sqrt{3};;vee;; 1+sqrt{2}\{9cdot3over4};;vee;; 1+2sqrt{2}+2\{27over4}-{12over4} ;;vee;; sqrt{8}\sqrt{225over16};;vee;; sqrt{128over16}RIghtarrow {3over2}sqrt{3}>1+sqrt{2}$

$displaystyle underset{D}max;{z}=zleft({piover3};{piover3}right)={3sqrt{3}over2}\ underset{D}min;{z}=zleft(0;0right)=0$

ОТВЕТ:

$displaystylelargeunderset{D}max;{z}=zleft({piover3};{piover3}right)={3sqrt{3}over2}\ underset{D}min;{z}=zleft(0;0right)=0$