Constraints And Lagrange Multipliers
Let us see the introduction about constraints of Lagrange multipliers
. In mathematical, the Lagrange multipliers method is used to provide a plan for finding the function of maxima and minima and its function subject to constraints. The Lagrange multipliers method of yields is very important condition for constrained problems. Let us see the definition of constraints and Lagrange multipliers.
Constraints and Lagrange Multipliers - Definition
The Lagrange multipliers function is defined as the, P(x, y, ?) = f(x, y) + ?.(g(x, y)-c).
Maximize f(x, y)
Subject to constraints g(x, y) =c
Let us see the example problems for Lagrange multipliers constraints.
Examples for Lagrange Multipliers Constraints:
Example 1:
Find the maximum and minimum of f(x, y) = 6x+4y subject to the constraint x2+y2=200
Solution:
Step 1:
We know that a minimum and maximum value,
Solve the given system,
6=2 ?x
4=2 ?y
x2+y2=200
Step 2:
We cant have ?=0, the first two equations are not satisfied.
X= 6/2?
Y=4/2?
Step 3:
The constraints gives,
36/4?2+16/4?2=200
52/4?2=13/?2=200
Step 4:
Solve for ?,
For ?= 0.254
Step 5:
For ?=0.254, 6=2(0.254) x
X=11.76
Y=4/2(0.254)
Y=7.87
Step 6:
For ?=-0.254,
X=-11.76
Y=-7.87
Step 7:
f (11.76, 7.87)=92.59
f(-11.74,-7.87)=-92.59
Step 8:
Answer is:
Maximum value at (11.76, 7.87)
Minimum value at (-11.76, -7.87)
Example 2:
Find the maximum and minimum of f(x, y) = 8x+6y subject to the constraint x2+y2=160
Solution:
Step 1:
We know that a minimum and maximum value,
Solve the given system,
8=2 ?x
6=2 ?y
x2+y2=160
Step 2:
We cant have ?=0, the first two equations are not satisfied.
X= 8/2?
Y=6/2?
Step 3:
The constraints gives,
64/4?2+36/4?2=160
100/4?2=25/?2=160
Step 4:
Solve for ?,
For ?= 0.395
Step 5:
For ?=0.395, 8=2(0.395) x
X=10.126
Y=6/2(0.395)
Y=7.594
Step 6:
For ?=-0.395,
X=-10.126
Y=-7.594
Step 7:
f (10.126, 7.594)=76.89
f(-10.126,-7.594)=-76.89
Step 8:
Answer is:
Maximum value at (10.126, 7.594)
Minimum value at (-10.126, -7.594)
These are the example problems for Lagrange multipliers constraints.
An interpolation is an estimation of the value of function whose values at near by pints on either side are known, rather than actually calculating it. In linear functions the estimation is exactly the actual value but in case of other functions it is closer to the actual depending on the closeness of the interval.
In some functions it may be difficult to easily find the value at a particular point. Lagrange interpolation helps us to determine the approximate value of the function by figuring out a different polynomial function for a particular interval in which the point lies. This function is always easier to work.
Let us study in detail in the next part of the topic.
Check this
substitution method examples awesome i recently used to see.
Basic Concept of Lagrange Interpolation
Lagrange interpolation
Look at the above graph. Let f(x) is the given function for which a set of values are known in the interval [1, 3] and we need to find its value at x = a. Now suppose a function g(x) has approximately the same values in the interval [1, 3], then it can be seen that it is easier to work with this function to find the value when x = a. The function g(x) is called Lagrange polynomial for that particular interval and the method to find the function is called Lagrange interpolation.
It may be clearly noted that the Lagrange polynomial can be used only for interpolation and not for extrapolation. For example, if you attempt to find the value of f(x) at x =0, the value of g(x) at x = 0, is way out from the actual value of f(x) at that point.
In general, let us consider let us a function f(x) and we know the set of function values in an interval [0, k] generally denoted as [xj, yj] in which no two xj are same.
The Lagrange interpolation is given by the following general polynomial,
$ L(j) = sum_{j=0}^{k}y_{j}l_{j}(x))$
in which lj (x) is given by,
$ l_{j}(x))=prod_{m=0}^{k}frac{x-x_{m}}{x_{j}-x_{m}}, m
eq j$
Explicitly, in the above product, the term is ignored when xm = xj
The Lagrange polynomial will be of n or less when the number of known set of values used is (n + 1).
The following worked out examples will show how Lagrange interpolation helps.
Example Problems
Below are the examples on lagrange interpolation -
Example 1:
Find the value of 2.54 by Lagrange interpolation.
Consider the obvious values of the function y = x4.
x0 = 1 y0 = 1
x1 = 2 y1 = 16
x2 = 3 y3 = 81
$l_{0}(x)=frac{x-x_{1}}{x_{0}-x_{1}}*frac{x-x_{2}}{x_{0}-x_{2}}= frac{x-2}{1-2}*frac{x-3}{1-3}$
= (1/2)(x2 5x + 6)
$l_{1}(x)=frac{x-x_{0}}{x_{1}-x_{0}}*frac{x-x_{2}}{x_{1}-x_{2}}= frac{x-1}{2-1}*frac{x-3}{2-3}$
= -(x2 4x + 3)
$l_{1}(x)=frac{x-x_{0}}{x_{2}-x_{0}}*frac{x-x_{1}}{x_{2}-x_{1}}= frac{x-1}{3-1}*frac{x-2}{3-2}$
= (1/2)(x2 3x + 2)
Therefore, L(x) = l0(x)y0 + l1(x)y1 + l2(x)y2
= (1/2)(x2 5x + 6)(1) - (x2 4x + 3)(16) + (1/2)(x2 3x + 2)(81)
= (1/2)[(x2 5x + 6) - (x2 4x + 3)(32) + (x2 3x + 2)(81)]
= 25x2 - 60x + 36
The approximate value of 2.54 is given by,
L(2.5) = 25(2.5)2 60(2.5) + 36 = 42.25
It may be noted that the exact value of 2.54 is 39.0625.
Example 2:
Find the value of sin 75o by Lagrange interpolation.
Consider the function y = sin(?x/6). Because we know the values of principal angles of sine function,
x0 = 1 y0 = 0.5
x1 = 2 y1 = 0.87
x2 = 3 y3 = 1
$l_{0}(x)=frac{x-x_{1}}{x_{0}-x_{1}}*frac{x-x_{2}}{x_{0}-x_{2}}= frac{x-2}{1-2}*frac{x-3}{1-3}$
= (1/2)(x2 5x + 6)
$l_{1}(x)=frac{x-x_{0}}{x_{1}-x_{0}}*frac{x-x_{2}}{x_{1}-x_{2}}= frac{x-1}{2-1}*frac{x-3}{2-3}$
= -(x2 4x + 3)
$l_{1}(x)=frac{x-x_{0}}{x_{2}-x_{0}}*frac{x-x_{1}}{x_{2}-x_{1}}= frac{x-1}{3-1}*frac{x-2}{3-2}$
= (1/2)(x2 3x + 2)
Therefore, L(x) = l0(x)y0 + l1(x)y1 + l2(x)y2
= (1/2)(x2 5x + 6)(0.5) - (x2 4x + 3)(0.87) + (1/2)(x2 3x + 2)(1)
= -0.12x2 + 0.73x 0.11
The approximate value of sin 75o is given by,
L(2.5) = -0.12(2.5)2 + 0.73(2.5) 0.11= 0.965
It may be noted that the exact value of sin 75o is 0.9659..
by: mathqa
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