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ejercicio calculo deifernecial

Module by: santiago cifuentes. E-mail the author

Summary: otimizacion

OPTIMIZACION

un comedero de caballo se debe hacer con las dimensiones que se muestran solamente se puede variar el angulo teta. que valor de teta maximizara el volumen del comedero.

el comedero es un prisma y las tapas del comedero son un trapecio (fugura con dos triangulos a los lados) el nagulo de el triangulo es teta y su longitud es x la hipoenusa de cada triangulo mide una pulgada y un lado y la base del trapecio mide una pulgada.

la longituda del comedero es de 20 pulgadas.

A trapecio = ( Bmayor + Bmenor 2 ) h A trapecio = ( Bmayor + Bmenor 2 ) h
(1)
Atrap = ( 2 x + 1 ) + ( 1 ) 2 Y Atrap = ( 2 x + 1 ) + ( 1 ) 2 Y
(2)
= ( 2 x + 2 2 ) y = ( 2 x + 2 2 ) y
(3)
= 2 ( x + 1 ) 2 y = 2 ( x + 1 ) 2 y
(4)
y 2 + x 2 = 1 y 2 = 1 - x 2 y = x + 1 2 y 2 + x 2 = 1 y 2 = 1 - x 2 y = x + 1 2
(5)
( x + 1 ) y = 0 ( x + 1 ) * 1 - x 2 = 0 ( x + 1 ) y = 0 ( x + 1 ) * 1 - x 2 = 0
(6)
d ' = 1 ( 1 - x 2 ) 1 2 ] + ( x + 1 ) ( 1 2 1 - x 2 ) ( - 2 x ) ] d ' = 1 ( 1 - x 2 ) 1 2 ] + ( x + 1 ) ( 1 2 1 - x 2 ) ( - 2 x ) ]
(7)
d ' = 1 - x 2 ] + [ - 2 x 2 - 2 x 2 ( 1 - x 2 ) d ' = 1 - x 2 ] + [ - 2 x 2 - 2 x 2 ( 1 - x 2 )
(8)
d ' = ( 1 - x 2 ) ( 2 ( 1 - x 2 ) ) + ( - 2 x 2 - 2 x ) 2 ( 1 - x 2 ) d ' = ( 1 - x 2 ) ( 2 ( 1 - x 2 ) ) + ( - 2 x 2 - 2 x ) 2 ( 1 - x 2 )
(9)
d ' = [ 2 - 2 x 2 - 2 x 2 - 2 x 2 1 - x 2 d ' = [ 2 - 2 x 2 - 2 x 2 - 2 x 2 1 - x 2
(10)
d ' = 2 ( 1 - 2 x 2 - x ) 2 ( 1 - x 2 ) d ' = 2 ( 1 - 2 x 2 - x ) 2 ( 1 - x 2 )
(11)
d ' = - 2 x 2 - x + 1 1 - x 2 d ' = - 2 x 2 - x + 1 1 - x 2
(12)
- b + - b 2 - 4 ac 2 a - b + - b 2 - 4 ac 2 a
(13)
- 1 + - 1 - 4 ( - 2 ) ( 1 ) 2 ( - 2 ) - 1 + - 1 - 4 ( - 2 ) ( 1 ) 2 ( - 2 )
(14)
- 1 + - 9 - 4 - 1 + - 9 - 4
(15)
- 1 + - 3 4 - 1 + - 3 4
(16)
- 1 + 3 4 = 1 2 = x - 1 + 3 4 = 1 2 = x
(17)
- 1 - 4 3 = - 1 = x - 1 - 4 3 = - 1 = x
(18)

se reemplaza en la funcion pitagoras del trapecio

x 2 + y 2 = 1 x 2 + y 2 = 1
(19)
y 2 = 1 - x 2 y 2 = 1 - x 2
(20)
y = 1 - ( 1 2 ) 2 y = 1 - ( 1 2 ) 2
(21)
y = 3 4 y = 3 4
(22)
y = 3 2 y = 3 2
(23)

se halla en angulo maximo

tan θ = 1 2 3 2 tan θ = 1 2 3 2
(24)
tan θ = 2 2 3 tan θ = 2 2 3
(25)
θ = tan - 1 1 3 θ = 30 θ = tan - 1 1 3 θ = 30
(26)

se halla el volumen de toda la figura

v = ( x + 1 ) y ( 20 ) v = ( x + 1 ) y ( 20 )
(27)
v = ( x + 1 ) 3 2 ( 20 ) v = ( x + 1 ) 3 2 ( 20 )
(28)
v = ( 1 2 ) + 1 ( 20 3 2 ) v = ( 1 2 ) + 1 ( 20 3 2 )
(29)
v = ( 3 2 ) ( 20 3 2 ) v = ( 3 2 ) ( 20 3 2 )
(30)
v = 60 3 2 v = 60 3 2
(31)
v = 15 3 v = 15 3
(32)
(33)

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