This document provides instructions to solve an optimization problem involving functions. The student is asked to find the value of the function f(t) by first determining the values of x, y, and z from a system of equations. This is done by taking the derivatives of the equations and setting them equal to solve for the variables. Once values for x, y, and z are obtained, they are substituted into the function f(t) and its derivative is evaluated to find the final value.
Operations Management - Book1.p - Dr. Abdulfatah A. Salem
Ejercicios melvis
1. REPÚBLICA BOLIVARIANA DE VENEZUELA
MINISTERIO DEL PODER POPULAR PARA
LA EDUCACIÓN UNIVERSITARIA
INSTITUTO UNIVERSITARIO POLITÉCNICO
“SANTIAGO MARIÑO”
EXTENSIÓN COL – CABIMAS
Alumno:
TSU Melvis Hernández
C.I 19.099.099
Cabimas, Mayo del 2016
OPTIMIZACION DE SISTEMAS Y FUNCIONES
2. EJERCICIOS
H A L L A R E L V A L O R D E L A F U N C I Ó N F ( T ) . P A R A E L L O ,
S E D E B E D E T E R M I N A R E L V A L O R D E X , Y Y Z
E M P L E A N D O E L M É T O D O Q U E S E I N D I C A ( E N
A L G U N O S C A S O S Q U E S E I N D I Q U E D E B E S D E R I V A R ) ,
L U E G O A P L I C A S L A R E S P E C T I V A D E R I V A D A E N L A
F U N C I Ó N F ( T ) Y S U S T I T U Y E S L O S V A L O R E S
O B T E N I D O D E X , Y Y Z Y R E A L I Z A S E L C Á L C U L O .
D E B E S E X P L I C A R C A D A P A S O . S E R E S O L V E R Á S O L O
U N E J E R C I C I O D E A C U E R D O A S U T E R M I N A L D E
C É D U L A . L O S V A L O R E S S E R E P R E S E N T A R A N E N
F R A C C I O N E S Y N O E N D E C I M A L E S
IGUALACIÓN
PARA CÉDULAS QUE TERMINEN EN 6, 7, 8 Y 9
-2X + 3Y – (Z2)´ = -1
X - 2Y – 3Z = -3
3
-X + 3Y + 2Z = 2
2
F(t) = 2Y” + Z4” – 3X2 ´
NOTA: ESTE SÍMBOLO ´ REPRESENTA LA PRIMERA DERIVADA Y “ A LA SEGUNDA
DERIVADA.
3. SE EXTRAE LA DERIVADA DE LA
ECUACIÓN I
−2𝑥 + 3𝑦 − 2𝑧 = −1
3𝑦 = −1 + 2𝑥 + 2𝑧
𝑦 =
−1 + 2𝑥 + 2𝑧
3
𝑥 − 6y − 9z = −9
−𝑥 +
3𝑦
2
+ 2z = 2
−2𝑥 + 3𝑦 + 4𝑧 = 4
Despejamos a Y
𝑥
3
−2𝑦−3𝑧=−3
III
I
II Ubicamos de forma
lineal las ecuaciones
4. El Sistema
Quedaría
−2𝑥 + 3𝑦 − 2𝑧 = −1
𝑥 − 6𝑦 − 9𝑧 = −9
−2𝑥 + 3𝑦 + 4𝑧 = 4
−2𝑥 + 3𝑦 + 4𝑧 = 4
3𝑦 = 4 − 4𝑧 + 2𝑥
𝑦 =
4 − 4𝑧 + 2𝑥
3
Despejamos a Y
Igualamos I/II
Y = Y
−1 + 2𝑥 + 2𝑧
3
=
4 − 4𝑧 + 2𝑥
3
Igualamo
s
−1 + 2𝑥 + 2𝑧 = 4 − 4𝑧 + 2𝑥
6𝑧 = 4 + 1
−1 = 4 − 6𝑧
𝑧 = 5
6
Eliminamos Términos Semejantes
Despejamos a Z
II
I