Методические указания к выполнению расчетно-графической и курсовой работы (1013347)
Текст из файла
(! "#$%%&' "$(%)*$ +)' %)#$! )"$")________________________________________.$ ! “"$0 ")*$ + 1 +)2$!%$")+ ”..55“6-655“””2008...1:1.2.3.4.5.6. +#,!"#...................... 2................................... 9................................... 14....................... 22.................................... 28$ ......................... 33$$$$:#$«"#........................
37»%&...21.! ": f (X) = 2x 2 + 2 y 2 + 2x y + 20x + 10 y + 10"extr:)$%(&'"&"!""%"'):4x + 2 y + 20f (x) =4 y + 2x + 10!/4x + 2 y + 20 = 04 y + 2x + 10 = 0#(1)(1) 2 ( 2)1fx24x + 20 = 0y=0x= 5y=0X* = ( 5, 0)$0$ :4x + 2 y + 20 = 06y = 0$2$:2f= 2;x y= 4;22f= 2;y xfy22 $4 22 4H( X ) == 4;1$$ :H(X*) =34 22 4$$ ,505:=4>02 = 4 4 2 2 = 12 > 0165,55-,,5,,$5 ,X* = ( 5, 0).«"#»%&1$...3)&0X = (0, 0)!!%&%'!)- ):#,!$)%!! $+%#%%$&! *! &!.(-1).0X0 =00f (X 0 ) = 2 0 2 + 2 0 2 + 2 0 0 + 20 0 + 10 0 + 10 = 10f (X 0 ) =4 0 + 2 0 + 2020=4 0 + 2 0 + 1010f (X 0 ) = 20 2 + 10 2 = 22.3607+12 $X1 =$00X120=100.1: X1 = X 0t 0 f (X 0 ) . !t 0 = 0.121f (X1 ) = 2 ( 2) 2 + 2 ( 1) 2 + 2 ( 2) ( 1) + 20 ( 2) + 10 ( 1) + 10 = 26f (X1 ) < f (X 0 ) ,f ( X1 ) =5,$4 ( 2) + 2 ( 1) + 2010=4 ( 1) + 2 ( 2) + 102f (X1 ) = 10 2 + 2 2 = 10.1980+22 $X2 =X210=0.12: X 2 = X1$21t 1 f ( X1 ) . !t 1 = 0.131.2f (X 2 ) = 2 ( 3) 2 + 2 ( 1.2) 2 + 2 ( 3) ( 1.2) + 20 ( 3) + 10 ( 1.2) + 10 = 33.92f (X 2 ) < f (X1 ) ,5,$«"#,(-1)»%&&, $.f (X 2 ) =..44 ( 3) + 2 ( 1.2) + 20=4 ( 1.2) + 2 ( 3) + 105.60.8f (X 2 ) = 5.6 2 + ( 0.8) 2 = 5.6569+32 $X35.60.1=0.8: X3 = X2$31.2X3 =t 2 f (X 2 ) .
!t 2 = 0.13.561.12f (X 3 ) = 2 ( 3.56) 2 + 2 ( 1.12) 2 + 2 ( 3.56) ( 1.12) + 20 ( 3.56) + 10 ( 1.12) + 10 = 36.5696f (X 3 ) < f (X 2 ) ,f (X 3 ) =5,$4 ( 3.56) + 2 ( 1.12) + 203.52=4 ( 1.12) + 2 ( 3.56) + 101.6f (X 3 ) = 3.52 2 + ( 1.6) 2 = 3.8666#$0123xyt0-2-3-3.560-1-1.2-1.120.10.10.1)&0X = (0, 0)0. ++1X1 =2 $$00t020=10102-0.8-1.6f|| f(X)||10-26-33.92-36.569622.360710.1985.65693.8666* !0X1y20105.63.52)'+2 $x%0-&%.: X1 = X 0t 0 f (X 0 ) .20 t 010 t 0t0:f (X1 ) = 2 ( 20 t 0 ) 2 + 2 ( 10 t 0 ) 2 + 2 ( 20 t 0 ) ( 10 t 0 ) + 20 ( 20 t 0 ) + 10 ( 10 t 0 ) + 10 == 800 t 02 + 200 t 02 + 400 t 02400 t 0«100 t 0 + 10 = 1400 t 02"#500 t 0 + 10»%&...5df (X1 )= 2800 t 0dt 0500 = 020 0.17857=10 0.17857X1 =f (X1 ) = 1400 0.17857 2f ( X1 ) =t0 =500= 0.1785728003.75141.7857500 0.17857 + 10 = 34.64294 ( 3.5714) + 2 ( 1.7857) + 20=4 ( 1.7857) + 2 ( 3.5714) + 102.14294.2857f (X1 ) = 2.1429 2 + ( 4.2857) 2 = 4.7916#$01xyt0-3.57140-1.78570.17857!)&X 0 = (0, 0)+12 $$00t010-4.2857f|| f(X)||10-34.642922.36074.7916--%0X10-%.: X1 = X 0[t0 f (X0 )].x1.20 t 020=00t0:f (X1 ) = 2 ( 20 t 0 ) 2 + 2 (0)2 + 2 ( 20 t 0 ) (0) + 20 ( 20 t 0 ) + 10 (0) + 10 == 800 t 02 400 t 0 + 10df ( X1 )= 1600 t 0 400 = 0dt 0X1 =&'0.
+X1 =y202.1429++2 $x20 0.25=0t0 =400= 0.25160050f ( X1 ) = 800 0.252 400 0.25 + 10 = 40«"#»%&.f ( X1 ) =..64 ( 5) + 2 (0) + 200=4 (0) + 2 ( 5) + 100f ( X1 ) = 02 + 02 = 06. .X1$f (X 2 ) =2 $$#0,0X1 -$!!$01xyt0-5000.25)&0X = (0, 0)y100f|| f(X)||10-4022.36070"!%0. +00-+1. +11-+2$X2.:X = X1 + t 1d 1d1 =0X2 =2 $=f (X1 ) +f ( X1 )2f (X 0 )20d0,d0 =f (X 0 )4.7916 2= 0.0459222.3607 22.1429d1 =+ 0.045924.28570%.2=&'+2 $x2003.5714+ t11.7857203.0612=103.82653.0612=3.82653.5714 3.0612 t 11.7857 + 3.8265 t 1t1:f (X 2 ) = 2 ( 3.5714 3.0612 t 1 ) 2 + 2 ( 1.7857 + 3.8265 t 1 ) 2 ++ 2 ( 3.5714 3.0612 t 1 ) ( 1.7857 + 3.8265 t 1 ) ++ 20 ( 3.5714 3.0612 t 1 ) + 10 ( 1.7857 + 3.8265 t 1 ) + 10 =«"#»%&...7= 2 (12.7548 + 21.8655 t 1 + 9.3709 t 12 ) + 2 (3.1887 13.6659 t 1 + 14.6421 t 12 ) ++ 2 (6.3774 13.6659 t 1 + 5.4663 t 1 11.7136 t 12 )71.428 61.224 t 1 17.857 + 38.265 t 1 + 10 == 25.5096 + 43.731 t 1 + 18.7418 t 12 + 6.3774 27.3318 t 1 + 29.2842 t 12 +23.4272 t 12+ 12.7548 16.3992 t 1= 24.5988t 1222.959 t 134.6432df (X 2 )= 49.1976 t 1 22.959 = 0dt 13.5714 3.0612 0.46666=X 2 ==1.7857 + 3.8265 0.46666f (X 2 ) = 24.5988 0.46666 2f (X 2 ) =71.428 61.224 t 1 17.857 + 38.265 t 1 + 10 =22.959= 0.4666649.19764.999935t1 =0.00003022.959 0.46666 34.6432 = 40.000324 ( 4.99993) + 2 ( 0.00003) + 200.0002=4 ( 0.00003) + 2 ( 4.99993) + 100.000024000f (X 2 ) = 0.0002 2 + 0.00002 2 = 0.0002 06.
.X2$#f (X 2 ) =0,0X2 -$!$012xytxy00-2010dxdy--20-10xyt-3.5714-1.78570.178572.1429dxdy0.04592-3.06123.8265xxyt-500.46666«y-4.2857x"0#|| f(X)||1022.3607f|| f(X)||-34.64294.7916fy0f-40»%&|| f(X)||0.)..8&).&)%&%X 0 = (0, 0)'+0. ++12 $00-X1$: X1 = X 02 $:4 22 4H(X 0 ) =.H 1 (X 0 ) =H 1 (X 0 ) f (X 0 )142112,24$$H 1 (X 0 ) =1316X0 :16136X1 =0013161613200=10020 3 10 60=20 6 + 10 305=050f (X1 ) = 2 ( 5) 2 + 2 0 2 + 2 ( 5) 0 + 20 ( 5) + 10 0 + 10 = 40f ( X1 ) =4 ( 5) + 2 0 + 200=4 0 + 2 ( 5) + 100f (X1 ) = 0 2 + 0 2 = 06. .$0,0f ( X1 ) =X1#X1 -$$01xyxy0-500200100«"#f|| f(X)||10-4022.36070»%&!...92.*: f (X) = x 12 + 2x 2 2 2x 12x 1 + x 2 = 1:6x 2#$&;C = const .
+•$•$25"=01 (X)= 2x 1 + x 2 + 1%5$1 (X%extrj (X)$Kacf = C,$,f ( X Kac ) =!2x 1 + x 2 + 1 = 0% !$12:2x 1 + x 2 = 1)0 *-.$2x 1):<' (:: 2x 1+ x2'$Kac24 x 2 Kac621=52x 1Kac:(= 1:22=4x 2Kac61,:2x 1 + x 2 = 1x 1 1 = 4x 22x 1 + x 2 = 164x 2 = 5x1"$ -f = ( 1) 2 + 2 12#$3::$$)f = 13 ,. . D > 0,/!x 12 + 2x 222x 16x 2x 12 + 2 x 222x16x 2 = 1x 122( x 22.#,:.$2x 1 +$:12 = 13#)x 2* = 1x 2 = 1 2x1 .:1 0=20 29x 2 = 9.$ -3D=X * = ( 1, 1) -$x 1* = 12 ( 1) 6 1 12 = 13 .:$(1) 2 (2)2x 1 + x 2 = 1,:3x 2 ) = 12«"#»%&...109 )= 1x 12 2 x 1 + 1 1 + 2( x 22 2 3 x 2 + 92 4444142431444234( x 1 1) 23 )221 + 2(( x 2( x 1 1) 2 + 2( x 29 )= 143 )2 = 1 + 1 + 9223 )2( x 1 1) 2 ( x 22 =1 +9924>/-$x1 = 1$x2 = 3"$(x 2/3 )22 =194: (1, 0)( x 1 1) 2=192$3 )2 = 924(x 2$22$x 1 = 1.12132(3.1213, 3 )2/,x1:x1 = ± 1(x 2x2$,3 )229 +1294x100.511.522.53#x2 = 0x 2 = 3.1213x1 1 = ± 3: ( 1.1213, 3 )25x2 = 33 =±322x2(1, 3)( x 1 1) 2 = 9$</x2 = 3 .2:: x1 = 1/"#/(1, 3 ) .2$1#$x112.581133.121332.58111$1-0.5811-1-1.1213-1-0.58111f = 13 .:«"#»%&$...10x2(X) = 0f=0f * = -131X*x110«"#»%&..)0 *.11&%(%1 !&'"&!$L(X, ) = x 12 + 2x 22!":%:?2x 112 +6x 2!%)"",, :1 (2x 1+ x 2 + 1)/$:L( X , )= 2x 1 2 + 2 1 = 0x1L( X , )= 4x 2x26+ 1 = 01 (X) = 2x 1 + x 2;$2x 1 2 + 26+4x 211::=02x 1 + 2=04x 2 +2x 1 + x 2 + 1 = 09111=2x2 + 2=6=6x2 =6,1–$3$?x 122=24 (1)+(2)(X * ,$<5:!L ( X, )=62L ( X, )=x1 x 2$/, :2L ( X, )=0x 2 x1L( X, )x 22=4d 2 L(X, ) = 2(dx 1 ) 2 + 4(dx 2 ) 2!$1 (X)x1=21 (X)x2*) = ( 1, 1, 2) -.$2=32x 1 + x 2 = 1$-11*1 =2x 2* = 1x1* = 141 x2x1 =22x 1 + x 2 = 14x 2 +(1) (3)=21612x 1 + x 2 = 1= 184x 2 ++1 = 01:=1d«1 (X)"= 2 dx 1 + 1 dx 2#»%&..2$..12X* = ( 1, 1, 2):d 2 L(X*) = 2(dx 1 ) 2 + 4(dx 2 ) 2d$ :dx 2 = 2dx 1d 2 L(X*) = 18(dx 1 ) 2 > 005,1 ( X*) = 2dx 1 + 1 dx 2 = 0 ,dx 1 ! 0X * = ( 1, 1)$$5.).*%;)%$,5:x2x 2 = 1 2x1 ,x2:::~f (X) = f ( x 1 ) == x 12 + 2( 1 2x 1 ) 2= 9x 12 + 18x 15~f (x1 ) :/~df (x1 )= 18x 1 + 18dx 1~d 2 f (x 1 )= 18 > 0d(x 1 ) 2"5$2x 1 + 6 12x 1 12 =4"36( 1 2x 1 ) 12 = x 12 + 2(1 + 4x 1 + 4 x 12 )2x 118x 1 + 18 = 0x1 * = 1~f (x1 )x 2 * : x 2 * = 1 2 ( 1) = 1$,x1* = 1$f (X)X * = ( 1, 1) .!) .*0%*:::F(X, r ) = x 12 + 2 x 222x 16x 212 +r(2 x 1 + x 2 + 1) 22!#&&,! &:F( X, r ) = f ( X )r m"2 j=12j (X)!:F(X, r )= 2x 1x12 + r (2x 1 + x 2 + 1) 2 = 0F(X, r )= 4x 2x26 + r (2 x 1 + x 2 + 1) = 0«"#»%&...13#:;$:( 2 + 4r ) x 1 + 2r x 2 = 2 2r2r x 1 + ( 4 + r ) x 2 = 6 r:5=2 + 4 r 2r= ( 2 + 4r ) ( 4 + r )2r4+r4r 2 = 8 + 16r + 2r + 4r 21=2 2r 2r= (26 r 4+r2r (6 r ) = 8 8r + 2r2=2 + 4r 2 2r= (2 + 4r) (6 r) 2r(2 2r) = 12 + 24r 2r 4r 22r6 r2r ) ( 4 + r )18r + 8,18r + 818r + 12*x 2 (r ) =18r + 8x 1* (r ) =62r 218r + 12=1x 2 * = limr # 18r + 8X * = ( 1, 1) -!$/= 2 + 4r > 02= (2 + 4r )(4 + r )0F(X, r )$ .: H ( X, r ) =11$:18r + 8= 1,x 1* = limr # 18r + 8$4r + 4r 2 = 18r + 12$/$12r + 2r 2 = 18r + 8./#2 + 4r 2r2r4+rr>054r 2 = 8 + 16r + 2r + 4r 2,:: 04r 2 = 8 + 18r > 05$,$f (X) .,$r>0$X = ( 1, 1) –*$5*1 :!*1= lim rr #= lim rr #:4r 2 = 18r + 8$"Ax1 , x 2218r + 8 18r + 12++ 1 = lim rr #18r + 818r + 836r + 16 + 18r + 12 + 18r + 8=18r + 836=218r + 8!#&:*j,= lim rr#«j ( X * ( r ))"#»%&...143.:*%f (X) = x 1 + x 2extr- x1 + x 2 $ 12 x1 + x 2 $ 4x1 , x 2 % 0!!(1)(2)(3):)0 *B•••&% !%$$:,,$Tf ( X ) = (1, 1)f (X) = C ,:/x1 + x 2 = 0 .$$(1)-(3);(0, 0) ;< :$$C : C = f (0, 0) = 0 + 0 = 0 ,(0, 0) .
B:x2ABx1CO1«"#»%&..E.155$$::,, , /A = (1, 2) . 6$,$.A$$:x 1*x *2=1=2f ( X *max ) = 1 + 2 = 35 $E,$, , /:). AO = (0, 0) . 6$$,$$:x 1*x *2=0=0f ( X *min ) = 0 + 0 = 0)#$5,$,.«"#»%&.)0 *..16&%.E"5$ :f (X) = x 1 + x 2max- x1 + x 2 $ 12 x1 + x 2 $ 4x1 , x 2 % 0##$$f (X) =x3, x 4 -:-$x 1 + x 2 + 0 x 3 + 0 x 4 max- x1 + x 2 + 1 x 3 + 0 x 4 = 12 x1 + x 2 + 0 x 3 + 1 x 4 = 4x1 , x 2 , x 3 , x 4 % 05212E$5::x2x3x41110015, .
.(2 2)).$$$$$x13:$5 2$ ,::(-:f (X) = x 1 + x 2 + 0 x 3 + 0 x 4max- x1 + x 2 + 1 x 3 + 0 x 4 = 12 x1 + x 2 + 0 x 3 + 1 x 4 = 4x1 , x 2 , x 3 , x 4 % 0E" $2$ :5x1 = 012-:x2 = 0x1 , x 2 /$$- x3 ,- x4x3 = 1x4 = 4$«"#(0, 0) .»%&...17H I @J KLFMD>@N@ ;411100CjCi99x1x2x3x4ri0x31-111010x54210141100Z-(; <=Z-(;<>?@AE,x3 =1x4 = 42 $1=12=16. .=x1 = 0x2 = 0>05 ,– Z-.
.5x2 . 0ri ,1r1 = = 11/r2 =5r1 ,$Z-/9x3 ,3 <•••/:Z-4=41+$:<,.$"'<DEEFAF@ ;@ @(G@;:2,22 $•1:1= 1 (0 + 0) = 121= 1 (0 + 0) = 1100001:<5:<– Z-,Z$. .:<.R = 1./:/:::22;x4 ;x2/2$:Z1-:Z2–Z;(11«"5$-1-111#:<&,/1100»,;%&)/15.•..18$$5:- 2-:/,:< :– /Z-(1),2:2-:02:<;1(1)*5;4413/$2-2-131101,01-110121100CjCi99x1x2x3x4ri10x2x413-13101-101-1120-10Z-(; <=Z-(;<>?@AE,x2 = 1x4 = 3:<x1 = 0x3 = 0x1 , x 2 /22 $$-(0, 1) .:11= 1 ( 1 + 0) = 20311= 0 (1 + 0) = 13 =0016. . 1 = max( 1 , 3 )1 >0,– Z12:=12 $$ri ,1= 1r1 =1x1 ,x4 ,"3 <•••$$Z-r2 ,Z-.
Характеристики
Тип файла PDF
PDF-формат наиболее широко используется для просмотра любого типа файлов на любом устройстве. В него можно сохранить документ, таблицы, презентацию, текст, чертежи, вычисления, графики и всё остальное, что можно показать на экране любого устройства. Именно его лучше всего использовать для печати.
Например, если Вам нужно распечатать чертёж из автокада, Вы сохраните чертёж на флешку, но будет ли автокад в пункте печати? А если будет, то нужная версия с нужными библиотеками? Именно для этого и нужен формат PDF - в нём точно будет показано верно вне зависимости от того, в какой программе создали PDF-файл и есть ли нужная программа для его просмотра.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
