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Taller 4: Optimización con Restricciones - Análisis Multivariado y Funcional, Ejercicios de Cálculo Avanzado

ejercicios practicos sobre variables y derivadas, integrales triples y gradientes

Tipo: Ejercicios

2019/2020

Subido el 14/09/2021

hpalenciaS
hpalenciaS 🇨🇴

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13/9/21 22:03
Taller 4-Hector Palencia
https://xlitemprod.pearsoncmg.com/api/v1/print/highered
1/2
Student: Hector Palencia
Date: 09/13/21
Instructor: Luis Eduardo Forero
Course: Análisis Multivariado y Funcional
(1)
Assignment: Taller 4
The temperature at point (x,y) on a metal plate is . An ant on the plate walks around the circle of
radius 5 centered at the origin. What are the highest and lowest temperatures encountered by the ant?
T(x,y) = 576x 576xy + 144y
2 2
First determine the equation for the circle centered at the origin with radius 5. This equation is shown below.
x y 25
2+2=
Suppose that f(x,y) and g(x,y) are differentiable. To find the local minimum of f subject to the constraint , find the values
of x, y, and that simultaneously satisfy the equations and . In this case,
and .
g(x,y) = 0
λ
f =
λ
g g(x,y) = 0
f(x,y) = T(x,y) = 576x 576xy + 144y
2 2 g(x,y) = x + y 25
2 2
Find .
f
f(x,y) = 576x 576xy + 144y
2 2
f = (1152x 576y) + (288y 576x)i j
Find .
g
g(x,y) = x y 25
2+2=
g = (2x)i (2y)j+
Apply the Lagrange method in order to find the values of r, h, and .
λ
f
= g
λ
(1152x 576y) + (288y 576x)i j = ( i j )
λ
2x + 2y
First solve the equation i for .(1152x 576y)i=
λ
2x
λ
(1152x 576y)i=i
λ
2x
576 288y
x=
λ
Now solve the equation j for .(288y 576x)j=
λ
2y
λ
(288y 576x)j=j
λ
2y
144 288x
y=
λ
Set the two expression for equal to each other and solve for one variable in terms of the other.
λ
576 288y
x=144 288x
y
576xy 288y2=144xy 288x2Multiply both sides by xy.
288x + 432xy 288y
2 2 = 0 Combine like terms on the left side.
144(2x y)(x + 2y) = 0 Factor.
Now set each factor equal to 0 and solve.
2x y = 0 or x 2y+ = 0
y = 2x x = 2y
Thus, the local extrema occur when y 2x or when x 2y.= =
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13/9/21 22:03 Taller 4-Hector Palencia

https://xlitemprod.pearsoncmg.com/api/v1/print/highered 1/

Student: Hector Palencia Date: 09/13/

Instructor: Luis Eduardo Forero Course: Análisis Multivariado y Funcional (1)

Assignment: Taller 4

The temperature at point (x,y) on a metal plate is. An ant on the plate walks around the circle of radius 5 centered at the origin. What are the highest and lowest temperatures encountered by the ant?

T(x,y) = 576x 2 − 576xy + 144y^2

First determine the equation for the circle centered at the origin with radius 5. This equation is shown below.

x 2 + y 2 = 25

Suppose that f(x,y) and g(x,y) are differentiable. To find the local minimum of f subject to the constraint , find the values of x, y, and that simultaneously satisfy the equations and. In this case, and.

g(x,y) = 0

λ ∇f = λ∇g g(x,y) = 0

f(x,y) = T(x,y) = 576x 2 − 576xy + 144y^2 g(x,y) = x 2 + y 2 − 25

Find ∇f.

f(x,y) = 576x 2 − 576xy + 144y^2

∇f = (1152x − 576y) + (288y − 576x) i j

Find ∇g.

g(x,y) = (^) x 2 + y 2 = 25

∇g = (2x) i +(2y) j

Apply the Lagrange method in order to find the values of r, h, and λ.

∇ f = λ∇g

(1152x − 576y) + (288y − 576x) i j = λ( 2x + 2y i j )

First solve the equation (1152x − 576y) i = λ2x i for λ.

(1152x − 576y) i = λ2x i

288y x

Now solve the equation (288y − 576x) j = λ2y j for λ.

(288y − 576x) j = λ2y j

288x y

Set the two expression for λequal to each other and solve for one variable in terms of the other.

288y x

= 144 − 288x y 576xy − 288y^2 = 144xy − 288x^2 Multiply both sides by xy. 288x 2 + 432xy − 288y^2 =^0 Combine like terms on the left side. 144(2x − y)(x + 2y) = 0 Factor.

Now set each factor equal to 0 and solve.

2x − y= 0 or x + 2y= 0 y = 2x x = −2y

Thus, the local extrema occur when y = 2x or when x = −2y.

13/9/21 22:03 Taller 4-Hector Palencia

  • https://xlitemprod.pearsoncmg.com/api/v1/print/highered 2/ - x 2 + y^2 = First consider y = 2x. Substitute y = 2x into the equation x 2 + y 2 = 25 to solve for x. - x 2 + (2x)^2 = - x = ±
    • So for this case, x = 5 and y = 2 5 or x = − 5 and y = − - x 2 + y^2 = Now consider x = − 2y. Substitute x = − 2y into the equation x 2 + y 2 = 25 to solve for y. - ( − 2y) 2 + y^2 = - y = ±
    • So for this case, y = 5 and x = − 2 5 or y = − 5 and x =
    • Evaluate T(x,y) for x = 5 and y = Substitute the values found for x and y into the equation for T(x,y) in order to find the minimum and maximum temperatures. - T(x,y) = 576x 2 − 576xy + 144y
      • T 5 ,2 5 = 576 5 2 − 576 5 2 5 + - =
    • Evaluate T(x,y) for x = − 5 and y = − - T(x,y) = 576x 2 − 576xy + 144y
      • T − 5 , − 2 5 = 576 − 5 2 − 576 − 5 − 2 5 + 144 − - =
    • Evaluate T(x,y) for x = − 2 5 and y = - T(x,y) = 576x 2 − 576xy + 144y
      • T − 2 5 , 5 = 576 − 2 5 2 − 576 − 2 5 5 + - = 18,
    • Evaluate T(x,y) for x = 2 5 and y = − - T(x,y) = 576x 2 − 576xy + 144y
      • T 2 5 , − 5 = 576 2 5 2 − 576 2 5 − 5 + 144 − - = 18,