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Algebra linial ejercicio, Apuntes de Álgebra Lineal

ejercicio de algebra linial ejercio muy eficaz

Tipo: Apuntes

2019/2020

Subido el 30/10/2020

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3.5 Constrained extrema and Lagrange mul-
tipliers
Example
Consider the ellipse x2+ 3y2= 1
We want to know at which points the distance d=px2+y2is min-
imal
Mathematically: We want to minimize px2+y2; at the same time the con-
dition x2+ 3y2= 1 should hold: the point Mis on the ellipse.
More generally, consider the functions: f:RnRand g:RnR.
The problem of constrained extrema is that of minimizing or maximizing
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3.5 Constrained extrema and Lagrange mul-

tipliers

Example

Consider the ellipse x^2 + 3y^2 = 1

We want to know at which points the distance d = √x^2 + y^2 is min- imal Mathematically: We want to minimize √x^2 + y^2 ; at the same time the con- dition x^2 + 3y^2 = 1 should hold: the point M is on the ellipse. More generally, consider the functions: f : Rn^ → R and g : Rn^ → R. The problem of constrained extrema is that of minimizing or maximizing

f (x) subject to the condition g(x) = 0. In our case f (x , y) = √x^2 + y^2 and g(x , y) = x^2 + 3y^2 − 1

“Lagrange multipliers”

Let f : Rn^ → R and g : Rn^ → R be C^1 functions. Define the set

S = {x ∈ Rn/g(x) = 0}.

Definition. We say that x 0 ∈ S is a global maximum for f subject to

the constraint g if f (x 0 ) ≥ f (x), ∀x ∈ S. We say that x 0 ∈ S is a global minimum for f subject to the constraint g if f (x 0 ) ≤ f (x), ∀x ∈ S.

Definition. We say that S is bounded (Spanish: acotado) if there exist

real constants x1 min, x1 max, · · · , xn min, xn max such that the following holds: for all x = (x 1 , · · · , xn) ∈ S we have x1 min ≤ x 1 ≤ x1 max, · · · , xn min ≤ xn ≤ xn max.

Theorem. Suppose that the following holds.

(i) S is bounded. (ii) ∀x ∈ S we have g′(x) 6 = ~0.

Then we have

(a) S contains at least a global maximum and a global minimum for f subject to the constraint g.

(2x 2 y) = λ (2x 6 y)

2 x = λ 2 x 2 y = λ 6 y

x(1 − λ) = 0 (1) y(1 − 3 λ) = 0 (2) If x = 0, then condition x^2 + 3y^2 = 1 gives ⇒ y = ± √^13 Substituting in (2) gives λ =^13 We have the critical points: A =^ ( 0 , √^13 ) and B =^ ( 0 , − √^13 ).

For A and B the value of λ =^13 If x 6 = 0, then λ = 1 because of (1). Substituting in (2) gives y = 0 The condition x^2 + 3y^2 − 1 = 0 gives x = ± 1 We have the critical points: C = (1 , 0) and D = (− 1 , 0). For C and D the value of λ = 1

To find the global minimum of the function f subject to the condition g we compute f (x) for all critical points x and we take the smallest value. In our case there are two global minima that correspond to the points A and B. To find the global maximum of the function f subject to the condition g we compute f (x) for all critical points x and we take the largest value. In our case there are two global maxima that correspond to the points C and D.