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Math Help - Solving Lagrange - optimal value of objective function

  1. #1
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    Solving Lagrange - optimal value of objective function

    I have the following problem:

    You have max(x2+y2) which his restrained on the set x2+y2=1
    1) Prove that this problem has solution.
    2) Solve the Lagrange problem
    3) Find the optimal value of the objective function

    Here my answers:
    1) It has a solution because the function is continous and it's restraint is a closed and bounded set. Therefore the extrem value theorem tells us that it attains a minimum and a maximum value at least once.

    2) L(x,y)=x2+y2*z(x2+y2-1)
    Now I take the three derivatives, set them zero and solve the system of equations, so I get the points:
    (-+1,0) and (0,+-1) and (+-21/2/2,+-21/2/2)
    So the first two points are the maximum while the third point is the minimum value.

    3) I don't really understand the question here. What could be meant by optimal value?
    I can see that the value of the function is 1 at its maximum and 1/2 at its minimum. - Could that be the answer?


    Thank you for your help!
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  2. #2
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    Re: Solving Lagrange - optimal value of objective function

    Are you sure that is the correct statement of the problem? With both the objective function and constraint function the same, x^2+ y^2, the objective function obviously has the constant value 1 at every feasible point.
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  3. #3
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    Re: Solving Lagrange - optimal value of objective function

    sorry, the function is:
    L(x,y)=x2*y2-z(x2+y2-1)

    for the calculations done in the first post I used the right function, it was a typing error.
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  4. #4
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    Re: Solving Lagrange - optimal value of objective function

    Quote Originally Posted by infernalmich View Post
    I have the following problem:

    You have max(x2+y2) which his restrained on the set x2+y2=1
    1) Prove that this problem has solution.
    2) Solve the Lagrange problem
    3) Find the optimal value of the objective function

    Here my answers:
    1) It has a solution because the function is continous and it's restraint is a closed and bounded set. Therefore the extrem value theorem tells us that it attains a minimum and a maximum value at least once.

    2) L(x,y)=x2+y2*z(x2+y2-1)
    Now I take the three derivatives, set them zero
    No, that is not how you solve a constrained min-max problem using Lagrange Multipliers- the points at which the partial derivatives are 0 may not satisfy the constraint. In order to find max or min of function f(x,y,z) subject to the constraint g(x,y,z)= constant, the gradients must be parallel. That is, \nabla f= \lambda\nabla g for some constant \lambda.

    and solve the system of equations, so I get the points:
    (-+1,0) and (0,+-1) and (+-21/2/2,+-21/2/2)
    So the first two points are the maximum while the third point is the minimum value.
    However, those points do satisfy the constraint. Did you take the partial derivatives of just the object function or a combination of the object function and the constraint? It is still not clear what the question was and what you did.

    3) I don't really understand the question here. What could be meant by optimal value?
    I can see that the value of the function is 1 at its maximum and 1/2 at its minimum. - Could that be the answer?


    Thank you for your help!
    "Optimal" just means "best" and what is "best" depends on the question. It really doesn't make sense here. I suspect that you are right- that they are referring to the actual max and min values at those points but "the optimal value" is awkward English.
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