How to solve using lagrange

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In summary, the conversation discusses solving a problem using Lagrange multipliers and includes three equations: dl/dx, dl/dy, and dl/dl. The goal is to find values for x, y, and backward l that minimize cost while satisfying the constraint x+y=2000. The conversation provides guidance on how to set the equations to zero and solve for the variables.
  • #1
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Homework Statement



if you have dl/dx= -2 +0.002x-lagrange function(backword L)

dl/dy=0.012y-5-lagrange function

dl/dl= -(x+y-2000)
How do you solve for x, y and backword l?

Homework Equations





The Attempt at a Solution

 
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  • #2
Is this problem about maximizing or minimizing a function subject to a constraint? If so, what are the function and constraint? I'm asking because I'm not convinced your partial derivatives are correct.

nikk834 said:

Homework Statement



if you have dl/dx= -2 +0.002x-lagrange function(backword L)

dl/dy=0.012y-5-lagrange function

dl/dl= -(x+y-2000)
How do you solve for x, y and backword l?

Homework Equations





The Attempt at a Solution

 
  • #3
Hi Vela , here is the problem that i am asked to solve.

The costs incurred by the company are denoted as follows:
R(x,y) = 7,500 -2x +0.006y^2-5y+0.001x^2

The company has a total of 2,000 knitting hours available each year, representing a constraint as follows:
x+y = 2,000

The company wishes to determine how to allocate the hours of operation between the two sweaters while minimizing cost.
(a)
Using the method of Lagrange, determine how many hours to allocate to each sweater.
(b)
Using the optimal hours determined above, what is the total cost?
 
  • #4
OK, so what I gather you did so far was to construct this function:

[tex]L(x,y,\lambda) = 7000-2x+0.006y^2-5y+0.001x^2-\lambda(x+y-2000)[/tex]

and took the partial derivatives. (The "backward L" is the Greek letter lambda.)

When you take the partial derivative with respect to x, for example, you have to also differentiate the term representing the constraint, so you'd get

[tex]\frac{\partial L}{\partial x}=-2+0.002x-\lambda[/tex]

Once you get the derivatives, to find the extrema, you set the partial derivatives to zero and solve the equations.
 
  • #5
Hi Vela,

I have done that so far and gotten 3 functions:

dl/dx=-2 + 0.002x-Langrange
dl/dy=0.012y-5-lagrange
dl/dl=-(x+y-2000)

I do not know how you would go about solving these equations but i know the solutions are x =1500 y=500 and lagrange=1

How exactly do you solve?
 
  • #6
Like I said, you set them equal to zero so you actually have equations to solve:

[tex]-2+0.002x-\lambda=0[/tex]

[tex]-5+0.012y-\lambda=0[/tex]

[tex]x+y-2000=0[/tex]
 
  • #7
If i solve it comes out to
0.002x=2L
this comes out as 1000 which is not right
and the next one comes out to 0.012y=5
this comes out to 416 which is not right
 
  • #8
You're not doing the algebra correctly. The first equation should become

[tex]0.002x = \lambda+2[/tex]

for example. Try solving for lambda in the first equation, and then substitute that into the second equation, so you just have x's and y's. Then solve for either x or y and substitute that into the last equation.
 
  • #9
ok so lamda=2-0.02x
if i substitude that into the next equation i get 0.012y-5-2-0.02x
how do i solve for that when i got 2 variables x and y.
 
  • #10
nikk834 said:
ok so lamda=2-0.02x
if i substitude that into the next equation i get 0.012y-5-2-0.02x
how do i solve for that when i got 2 variables x and y.
You can't solve that since, for one, it's not an equation. What does that expression equal? Once you have that sorted out, remember that you still have one more equation you can use to help you solve for x and y.
 

What is Lagrange's method and when is it used?

Lagrange's method, also known as the method of undetermined multipliers, is a mathematical technique used to find the maximum or minimum value of a function subject to a set of constraints. It is used in optimization problems where the objective function and constraints are defined by a set of equations.

What is the Lagrangian function and how is it used in Lagrange's method?

The Lagrangian function is a mathematical function that combines the objective function and constraints into a single equation. It is used in Lagrange's method by introducing a set of multipliers, known as Lagrange multipliers, to the Lagrangian function to form a new function that can be optimized.

What are the steps to solve a problem using Lagrange's method?

The steps to solve a problem using Lagrange's method are as follows:
1. Formulate the objective function and constraints as a set of equations.
2. Construct the Lagrangian function by combining the objective function and constraints.
3. Take the partial derivatives of the Lagrangian function with respect to each variable.
4. Set the derivatives equal to zero and solve for the variables.
5. Plug the values of the variables into the objective function to find the maximum or minimum value.

What types of problems can be solved using Lagrange's method?

Lagrange's method can be used to solve problems with multiple variables and constraints, such as optimization problems in economics, physics, engineering, and other fields. It can also be applied to problems in calculus, such as finding the extreme values of a function.

Are there any limitations to Lagrange's method?

Lagrange's method may not always provide a solution to a problem, especially when the objective function and constraints are highly nonlinear or when there are many variables and constraints involved. It also requires the existence of continuous partial derivatives of the objective function and constraints, which may not always be the case.

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