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optimizacion

Module by: diego diaz torres. E-mail the author

Summary: optimizacion

Encuentre las dimensions del cilindro circular recto mas grande que se puede inscribir en una esfera e radio 10cm. Encuentre las dimensions del cilindro circular recto mas grande que se puede inscribir en una esfera e radio 10cm.
(1)
V = π r 2 h h 2 = a 2 + b 2 V = π r 2 h h 2 = a 2 + b 2
(2)
100 = h 2 2 + r c 2 r c = 100 - h 2 4 100 = h 2 2 + r c 2 r c = 100 - h 2 4
(3)
V c = π 100 - h 2 4 2 h V c = π ( 100 - h 2 4 ) h = π h ( 100 - h 2 4 ) = 100 π - π h 3 4 V c = π 100 - h 2 4 2 h V c = π ( 100 - h 2 4 ) h = π h ( 100 - h 2 4 ) = 100 π - π h 3 4
(4)
V ´ ( h ) = 100 π - ( 3 π h 2 4 ) - ( π h 3 ) ( 0 ) 16 = 100 π - 12 π h 2 16 = 0 - 12 π h 2 16 = - 100 π - 12 π h 2 = - 100 π · 16 - 12 π h 2 = - 1600 π h 2 = - 1600 π - 12 π h 2 = 133 . 33 h = 133 . 33 = 11 . 54 V ´ ( h ) = 100 π - ( 3 π h 2 4 ) - ( π h 3 ) ( 0 ) 16 = 100 π - 12 π h 2 16 = 0 - 12 π h 2 16 = - 100 π - 12 π h 2 = - 100 π · 16 - 12 π h 2 = - 1600 π h 2 = - 1600 π - 12 π h 2 = 133 . 33 h = 133 . 33 = 11 . 54
(5)
r c = 100 - 133 . 33 2 4 = 100 - 133 . 33 4 = 100 - 33 . 25 = 66 . 75 = 8 . 17 r c = 100 - 133 . 33 2 4 = 100 - 133 . 33 4 = 100 - 33 . 25 = 66 . 75 = 8 . 17
(6)
V = π r 2 h , V = π 8 . 17 2 · 11 . 55 V = π 66 . 75 · 11 . 55 = 770 . 96 π cm 3 V = π r 2 h , V = π 8 . 17 2 · 11 . 55 V = π 66 . 75 · 11 . 55 = 770 . 96 π cm 3
(7)

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