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Ejercicios de Optimizacion 5

Module by: DIEGO DIAZ LEON. E-mail the author

Summary: Ejercicios de Optimizacion

OPTIMIZACION 5

1. un trozo de alambre de 10 m de largo se corta en dos partes. una se dobla para formar un cuadrado y la otra para formar un triangulo equilatero. hallar como debe cortarse el alambre de modo que el area encerrada sea:

a. maxima

b. minima

La parte x del alambre se usa para el cuadrado, por lo tanto cada lado tiene una longitud de x/4

La parte 10-x de alambre se usa para el triangulo equilatero, por lo tanto cada lado tiene longitud 1-x/3.

El dominio de x es (0,10)

h 2 + ( 10 - x 6 ) 2 = ( 10 - x 3 ) 2 ⇒ h 2 = 1 9 ( 10 - x ) 2 - 1 36 ( 10 - x ) 2 ⇒ h 2 = 3 36 ( 10 - x ) 2 ⇒ h = 3 6 ( 10 - x ) h 2 + ( 10 - x 6 ) 2 = ( 10 - x 3 ) 2 ⇒ h 2 = 1 9 ( 10 - x ) 2 - 1 36 ( 10 - x ) 2 ⇒ h 2 = 3 36 ( 10 - x ) 2 ⇒ h = 3 6 ( 10 - x )
(1)

el area del cuadrado es

Ac ( x ) = x 2 16 Ac ( x ) = x 2 16
(2)

area del triangulo es

At ( x ) = 1 2 * b * h ⇒ 1 2 * 1 3 ( 10 - x ) * 3 6 ( 10 - x ) ⇒ 3 36 ( 10 - x ) 2 At ( x ) = 1 2 * b * h ⇒ 1 2 * 1 3 ( 10 - x ) * 3 6 ( 10 - x ) ⇒ 3 36 ( 10 - x ) 2
(3)

Area de ambas figuras es

A ( x ) = At ( x ) + Ac ( x ) = 3 36 ( 10 - x ) 2 + x 2 16 A ( x ) = At ( x ) + Ac ( x ) = 3 36 ( 10 - x ) 2 + x 2 16
(4)
A ' ( x ) = 1 8 x + 3 36 * 2 ( 10 - x ) ( - 1 ) = 1 8 x - 3 18 ( 10 - x ) = 1 8 x - 5 3 9 + 3 18 x = 9 + 4 3 72 x - 5 3 9 A ' ( x ) = 1 8 x + 3 36 * 2 ( 10 - x ) ( - 1 ) = 1 8 x - 3 18 ( 10 - x ) = 1 8 x - 5 3 9 + 3 18 x = 9 + 4 3 72 x - 5 3 9
(5)

punto critico:

A ' ( x ) = 0 ⇒ 9 + 4 3 72 x - 5 3 9 = 0 ⇒ x = 72 ( 5 3 ) 9 ( 9 + 4 3 ) = 40 3 9 + 4 3 ≈ 4 . 3496 A ' ( x ) = 0 ⇒ 9 + 4 3 72 x - 5 3 9 = 0 ⇒ x = 72 ( 5 3 ) 9 ( 9 + 4 3 ) = 40 3 9 + 4 3 ≈ 4 . 3496
(6)
A ' ' ( x ) = 9 + 4 3 72 > 0 A ' ' ( x ) = 9 + 4 3 72 > 0
(7)

ahora se calcula en A(x) el dominio de la funcion.

A ( 0 ) = 3 36 100 ≈ 4 . 8112 & A ( 10 ) = 100 16 = 6 . 25 A ( 0 ) = 3 36 100 ≈ 4 . 8112 & A ( 10 ) = 100 16 = 6 . 25
(8)

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