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Calc Vectorial UnidaGd 4 B
Calc Vectorial UnidaGd 4 B
Calc Vectorial UnidaGd 4 B
DERIVADAS DIRECCIONALES
Si f es una funcin diferenciable de x e y, entonces la derivada direccional
de f en la direccin del vector unitario.
= cos + sin
Es:
(, ) = (, ) cos + (, ) sin
Ejemplo:
Calcular la Derivada direccional de f(x, y)=4-x^2-1/4y^2 en (1, 2) en la
= cos /3 + sin /3
direccin
1
(, ) = 4 2 2
4
= cos + sin
3
3
1
(, ) = (2) cos 60 + ( ) sin 60
2
(1,2) = 2(1) + (0.5)(2)(0.8660) = 1 0.8660 = 1.8660
6
8
6
(1, 4) = (1)(1) (1)2 (1) = = 1.2
5
5
5
GRADIENTE
Si z=f(x, y), entonces el gradiente de f1 que se denota mediante f(x, y)
es el vector
(, ) = (, ) + (, )
Otra denotacin
grad f (x, y)
(, )
Ejemplo:
Calcular la derivada direccional de (, ) = 3 2 2 2 en (-1,3) en la
direccin que va de P= (-1,3) a Q= (1, -2)
(, ) = 6 4 = (6 4)
(1,3) = [6(1), 4(3)] = (6, 12)
=
= = (1, 2) (1,3) = (2, 5)
|||| = (2)2 + (5)2 = 4 + 25 = 29
1
1
2
5
(2, 5) = (
=
=
,
)
||||
29
29 29
2
5
(1,3) = (1,3)
= (6, 12) (
,
)
29 29
12
60
48
=
+
=
= 8.9134
29 29 29
Ejemplo:
Calcular (, , ) para la funcin (, , ) = 2 + 2 4 de
crecimiento mximo de f en el punto (2, -1, -1)
(, , ) = 2 + 2 4 = (2, 2, 4) gradiente
(2, 1, 1) = [2(2), 2(1), 4] = (4, 2, 4)
Direccin de crecimiento mximo de f
= (1,2)
= (1,1)
b) (, ) =
= 1 + 3
Procedimiento
) (, ) = 3 4 + 5
= (1,2)
= 1 + 3
(, ) = (3 4) + (4 + 5) = (3 4, 4 + 5)
(1,2) = {[3 4(2)], [4(1) + 5]} = (3 8, 4 + 5) = (5,1)
1 2
2
1 3
13
= ( ) + (3) = + =
2
4 1
4
=
1
2 1
2 2 3
1 2 3
( , 3 ) =
( , 3) = (
)=(
)
,
,
213 13
13 2
13 2
13 13
2
1 2 3
5
2 3
)=
(1,2) = (5,1) (
,
+
13 13
13 13
= 1.3867 + 0.9607 = .
) (, ) =
= (1,1)
(, ) = 1
1
(, ) = 1 + (1) 2 = ( , 2 )
1 1
(1,1) = ( , 2 ) = (1, 1)
1 1
= (0, 1)
= 02 + (1)2 = 1
= (1, 1) (0, 1) = 0 + 1 =
(1,1) = (1,1)
= cos +
Hllese la derivada direccional de la funcin en la direccin
sin de:
) (, ) =
) (, ) =
2
3
Procedimiento:
) (, ) =
= [( + )2 + ( + )1 (1)] = [
1
+ +
]
[
]
+
=
( + )2 +
( + )2
(, ) = (, ) cos + (, ) sin
] sin 330
= ( + )2 cos 330 + [
( + )2
0.866
0.866
0.5
(
)
[
]
=
+
0.5
=
( + )2
( + )2
( + )2 ( + )2
0.866 0.5
=
( + )2
) (, ) =
2
3
a) (, ) = cos( + )
= (0, )
b) (, ) =
= (2,4,0) = (0,0,0)
= (2 , 0)
Procedimiento:
) (, ) = cos( + )
= sin( + )
= (0, )
= ( 2 , 0)
= sin( + )
(, ) = [ sin( + )] + [ sin( + )]
(0, ) = [ sin( ) sin()] = (0,0) = 0
(0, ) = (0, ) = 0
=
= = (2 , 0) (0, ) = ( 2 , )
) (, ) =
=
= (2,4,0) = (0,0,0)
(, , ) = ( , , )
(2,4,0) = (4 0 , 2 0 , 8 0 ) = (4,6,8)
=
= = (0,0,0) (2,4,0) = (2, 4,0)
=
(2,4,0) = (
20
2
,2
, 0) = (2
2
5
, 0) = (
1 2
(2,4,0) = (2,4,0)
= (4,2,8) (
5 5
2
5
, 0)
, 0) = (
4
5
4
5
+ 0) =
(4,2)
) = 2 2
(2,1,1)
Procedimiento:
1
(, ) = [( ) ()()2 ] + = [
2
(2,4) = (2
, ]
, 4) = (2 , 2)
4
Valor mximo (, ) = (, )
2
1
1
17
= (2) + (2)2 = 4 + 4 = 4 = 2.0616
b) (, , ) = [ 2 2 ] + [2 2 ] + [2 2 ] = ( 2 2 , 2 2 , 2 2 )
(2,1,1) = [(1)2 (1)2 , 2(2)(1)(1)2 , 2(2)(1)2 (1)] = (1,4,4)
( x ,y)= 2 6
4 = 8
2 = 6
=2
=3
EJEMPLO ANTERIOR
(, ) = 4
(, ) = 2
(, ) = 0
= ()() =
(2,3) = 2(2)2 + (3)2 + 8(2) 6(3) + 20 = Mnimo relativo
4
4
=
Sustituimos x en 1
3 2 + 4 = 0
(3 + 4) = 0
=
0
3 + 4
1 = 0
Puntos crticos:
4 4
(0,0)( , )
3 3
(, ) = 6
(, ) = 4
(, ) = 4
(0,0)
= [6(0)][4] [4]2 = 0 16 = 16
(0,0) =
16 4 = 4
8 2 = 2
2 + 4 = 10
14x
8 2 = 2
=14
2 = 2 8
X=1
y=3
(, ) = 8 > 0
(, ) = 4
(, )(, ) = 2
= (8)(4) (2)2
= 32 4 = 28 > 0
(1,3) = = 4(1)2 + 2(3)2 2(1)(3) 10(3) 2(1)
= 4 + 18 6 30 2 = 16
(, ) = 3 + 3 3 2 3 2 9---1
(, ) = 3 2 6 = 0---2
0
3( 2)
=
2=
0
3
2 =0
=
3( 2 2 3) = 0
2 2 3 = 0
( 3)( + 1) = 0
=
(, ) = 6 6
(, ) = 6 6
(, ) = 0
Puntos
crticos
(3,0)
(3,2)
(1,0)
(1,2)
(, )
(, )
33 + 03 3(3)2 302
9(3)
(12)(0 6) 02 = 72
6(3) 6 = 12
= 27 27 27
= 27
33 + 23 3(3)2
(12)(12 6) = 72 > 0 3(2)2 9(3)
6(3) 6 = 12
= 31 .
(1)3 + 03 3(1)2
3(0)2 9(1)
(12)(0 6) = 72
6(1) 6 = 12
= 13+9
= 5 .
(1)3 + 23 3(1)2
6(1) 6 = 12 (12)(12 6) = 72
3(2)2 9(1)
= 1
MULTIPLICACIN DE LAGRANGE
Para evaluar los extremos de = (, ) sujeto a la restriccin
(, ) = 0, resuelve el sistema de ecuaciones.
(, ) = (, )
(, ) = (, )
Ejemplo.
Aplicar el mtodo de multiplicacin de LaGrange para obtener el mximo.
(, ) = 9 2 2
+ = 3
(, ) = 2 = (1)--------1
(, ) = 2 = (1)--------2
+ =33
2
2 = 2
=
2
2
3
+ =3
+ =3
2 = 3
=
3
=
2
Mximo:
32 32
(, ) = 9
=
2
2
=2 4 4 =
3618
4
18
4
*DETERMINA EL MXIMO
f(x,y,z)= 2 + 2 + 2 sujeta a 2x-2y-z=5
2x=(2)..1
2y= (-2)2
2z= (-1)3
2x-2y-z=5..4
Despejar en ecuacin 1
2
=2 =
Sustituir en ecuacin 2 y ecuacin 3
2y=-2x2
2
y= 2 =
2z = -x.3
z= 2
Sustituir y y z en ecuacin 4
2x-2y-z=5
2x-2(-x)-(-2)=5
2x+2x+ =5
2
8x+x=10
9x=10
10
x= 9
10
y = -x = - 9
1 10
z= 2 = -2( 9 )= 9
10
mximo f( 9 ,
10
10 5
10 2
)
9
= ( 9 )2 + (
+ ( 9 )2 =
100
81
100
81
25
+ 81 =
225
81
= 2.7778
2 2
) (2) + 2 ] 2
2
44
4()
2 = 22=2() = 2
Sustituir 1 2 3
2z= 1 (2z)+ 2
2z=0+2
2
z= 2=1
Sustituir z en la ecuacin 4 y 5
2 + 2 + 1 = 11 2 + 2 = 10 ..4
x+y+1=3 x+y=2..5
x=2-y
Sustituir x en la ecuacin 4
2 + 2 = 10
(2 )2+ 2 =10
4-4y+ 2 + 2 = 10
2 2 4 + 4 10 = 0
2 2 4 6 = 0
2( 2 2 3)=0
2 2 3 =
(y-3)(y+1)=0
1 = 3
0
=0
2
2 = 1
x=2-y
1 = 2 3 = 1
2 = 2 (1) = 2 + 1 = 3
(-1,3) (3,-1)
(-1,3,1) (3,-1,1)
Despejar 2 de la ecuacin 1
-1 (2x)= 2
Sustituir 2 en ecuacin 2
2=-1 (2y)+21 (2)
2-2=-1 (2) 1 (2)
0=2-1 ( )
0
y-x= 2
y-x=0
y=x
2 4
2
6364(3)(1)
x=
2(3)
636+12
6
x=
x=
2(323)
2(3)
6(16)(3)
6
643
6
1 =
3 + 2 3
3
2 =
3 2 3
3
z= 3-2x
1 = 3 2 [
1 =
9 6 4 3
3
343
1 =
3 + 2 3
]
3
=1-
2 = 3 2 [
4 3
3
3 2 3
9 6 + 4 3
]=
3
3
3+43
2 =
(, ) = 100 4 4
Donde X representan las unidades de trabajo (a $ 150 por unidad). e Y las
unidades de trabajo de capital (a $ 250 por unidad).
El costo total de trabajo y capital est limitado a $ 50 000. Calcula el nivel
mximo de produccin para este fabricante.
Y4
1
2Y 4
Sustitucin en ecu. 2
1
100 3 3
4
4 4 = ( 1 ) 250
4
2 4
3
4
1
4
=
1
5 4
2
3
4
X = 5 4 4
X = 5Y
Sus X en ecu. 3
150 (5Y) +250 Y = 50 000
750 Y +250 Y =50 000
1000 Y =50 000
Y=
50 000
1000
Y= 50
X=5 Y
X= 5 (50)
X= 250
Y4
1
2X 4
504
=
2
1
(250)4
= 0.3344
Produccin mxima
3