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# Ejercicios de Optimizacion4

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

Summary: Ejercicios de Optimizacion

OPTIMIZACION4

1. Se desea construir un recipiente cilindrico de metal con tapa que tenga una superficie total de 80 cm cuadrados. determine sus dimensiones de modo que tenga el mayor volumen posible.

v = π r 2 h v = π r 2 h
(1)
A = 2 π r 2 + 2 π rh = 80 cm 2 A = 2 π r 2 + 2 π rh = 80 cm 2
(2)

una funcion

v = π r 2 h v = π r 2 h
(3)

una ecuacion

2 π r 2 + 2 π rh = 80 2 π r 2 + 2 π rh = 80
(4)
2 π r 2 + 2 π rh = 80 π r 2 + π rh = 40 π rh = 40 - π r 2 h = 40 - π r 2 π r 2 π r 2 + 2 π rh = 80 π r 2 + π rh = 40 π rh = 40 - π r 2 h = 40 - π r 2 π r
(5)
v = π r 2 h π r 2 ( 40 - π r 2 π r ) r ( 40 - π r 2 ) v = 40 r - π r 3 v = π r 2 h π r 2 ( 40 - π r 2 π r ) r ( 40 - π r 2 ) v = 40 r - π r 3
(6)
v ' = 40 - 3 π r 2 v ' = 40 - 3 π r 2
(7)
v ' = 0 40 - 3 π r 2 = 0 r 2 = 40 3 π 4 . 2441 v ' = 0 40 - 3 π r 2 = 0 r 2 = 40 3 π 4 . 2441
(8)
r = 4 . 2441 ± 2 . 0601 r = 4 . 2441 ± 2 . 0601
(9)

se toma el valor positivo de la raiz.

v ' = 40 - 3 π r 2 v ' ' = - 6 π r v ' = 40 - 3 π r 2 v ' ' = - 6 π r
(10)
v ' ' = - 6 π r - 6 π ( 2 . 0601 ) < 0 v ' ' = - 6 π r - 6 π ( 2 . 0601 ) < 0
(11)
h = 40 - π r 2 π r 40 - π ( 2 . 0601 ) 2 π ( 2 . 0601 ) 4 . 1203 h = 40 - π r 2 π r 40 - π ( 2 . 0601 ) 2 π ( 2 . 0601 ) 4 . 1203
(12)

las dimensiones del cilindro son:

r 2 . 0601 & h 4 . 1203 r 2 . 0601 & h 4 . 1203
(13)

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