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Thread: Question about Lagrange multipliers for maximizing a function with two constraints

  1. #1
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    Question about Lagrange multipliers for maximizing a function with two constraints

    Hi everyone.

    I'm not that familiar with English math terminology so I hope that you'll bear with me.

    Currently, I'm trying to maximize a function with two constraints, but I got stuck because of one of my constraints. My first constraint has both the variables x and y, but my second constraint only has the variabley. The reason why I'm confused by this is that when I proceed to solve the problem, I have no use for the Lagrange multiplier λ. I can simply solve the partial differential for λ1 and λ2. This will enough to yield my results (the x and y coordinates). It is frustrating me because I need to put it into words, what I am doing (in terms of using Lagrange multipliers) and why I apparently had to skip the λ all together.

    The function that I'm trying to maximize is as follows:
    𝑓(𝑥, &#119910 = −0,01𝑥2 + 395𝑥 + 100

    My constraints are these:
    2𝑥 + 𝑦 ≤ 44,000
    𝑦 ≤ 20,000

    I know that the correct answer (through using other methods) is:
    x = 12,000
    y = 20,000

    The way that I've proceeded to solve this problem is by putting the respective functions and constraints into an algorithm:
    L(x ,y, λ1, λ2) = −0,01𝑥2 + 395𝑥 + 100 - λ1 * (2𝑥 + 𝑦 - 44,000) - λ2 * (𝑦 - 20,000)

    Then finding the partial differentials of x, y, λ1 and λ2 to ultimately isolate x and y. I am getting the correct results, but there's no need to find the partial differentials of x and y?
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    Re: Question about Lagrange multipliers for maximizing a function with two constraint

    $2 + \le 44000$ means what? Is there supposed to be a variable in there?

    $\le 20000$ does not mean anything. There is nothing on the left hand side of the inequation.

    If you have two constraints, you will have $\lambda_1$ and $\lambda_2$ And no $\lambda.$

    But of course you are not multiplying your lambdas by anything except a constant so they have no effect on anything.

    Please indicate what the constraints are, and we can proceed from there.
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    Re: Question about Lagrange multipliers for maximizing a function with two constraint

    The original post does not seem to have been edited but I see " 2x+ y\le 44000" and " y\le 20000".
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    Re: Question about Lagrange multipliers for maximizing a function with two constraint

    Quote Originally Posted by HallsofIvy View Post
    The original post does not seem to have been edited but I see " 2x+ y\le 44000" and " y\le 20000".
    Actually I do not even see a function with variables.
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    Re: Question about Lagrange multipliers for maximizing a function with two constraint

    I see (minus some strange characters) the object function f(x, y)= -0.01x^2+ 395x+ 100 which, oddly, is a function of x only!

    johndoe69, differentiating the object function minus \lamda_1 and \lambda_2 times the constraints with respect to \lambda_1 and \lambda_2 just give the constraints again which does not really help- you already know them. It is the derivatives with respect to x and y that are crucial! I do not understand why you say "there's no need to find the partial derivatives o x and y" (I assume you mean "with respect to x and y).
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    Re: Question about Lagrange multipliers for maximizing a function with two constraint

    Yes, I see that there's been some problems with the formatting and some "spelling" issues when writing the equations. I'm not used to using LaTeX, but I'll try to rewrite some of the confusing things. Is there a way to edit/delete my original post, or do I have to write another reply to this thread?

    EDIT: For now, I suppose I'll add it to the rest of this reply:


    Hi everyone.

    I'm not that familiar with English math terminology so I hope that you'll bear with me.

    Currently, I'm trying to maximize a function with two constraints, but I got stuck because of one of my constraints. My first constraint has both the variables x and y, but my second constraint only has the variabley. The reason why I'm confused by this is that when I proceed to solve the problem, I have no use for the Lagrange multiplier \lambda. I can simply solve L_{\lambda_{1}} = 0 and L_{\lambda_{2}} = 0. This will enough to yield my results (the x and y coordinates). It is frustrating me because I need to put it into words, what I am doing (in terms of using Lagrange multipliers) and why I apparently had to skip the \lambda all together.

    The function that I'm trying to maximize is as follows:

    f(x,y) = -0,01x^2 + 395x + 100y


    My constraints are these:

     2x + y \leq 44,000

    y \leq 20,000


    I know that the correct answer (through using other methods) is:

    x = 12,000

    y = 20,000


    The way that I've proceeded to solve this problem is by putting the respective functions and constraints into an algorithm:

    L(x, y, \lambda_{1}, \lambda_{2}) = -0,01x^2 + 395x + 100y - \lambda_{1} * (2x + y - 44,000) - \lambda_{2} * (y - 20,000)

    Then figuring out the partial differentials with respect to x, y, \lambda_{1} and \lambda_{2} to ultimately isolate \lambda_{1} and \lambda_{2}. I am getting the correct results, but there's no need to solve L_x = 0 and L_y = 0?


    My partial derivatives are as follows:

    L_x = -0.02x + 395 - 2\lambda_{1}

    L_y = 100 - \lambda_{1}

    L_{\lambda_{1}} = -2x - y + 44,000

    L_{\lambda_{2}} = -y + 20,000
    Last edited by johndoe69; Nov 18th 2017 at 01:15 PM.
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    Re: Question about Lagrange multipliers for maximizing a function with two constraint

    You've set $L$ up correctly

    $\nabla L = \left(-2 \lambda _1-\dfrac{x}{50}+395,-\lambda _1-\lambda _2+100,-2 x-y+44000,20000-y\right)$

    and each term must be set to 0 and the system solved .

    The partials with respect to the $\lambda$'s are just the constraint conditions.

    This system is easily solved as

    $(x,y,\lambda_1, \lambda_2) = \left( 12000,20000,\dfrac{155}{2}, \dfrac{45}{2} \right)$
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    Re: Question about Lagrange multipliers for maximizing a function with two constraint

    The method of Lagrange multipliers (with two constraints)

    max or min f(x,y) subject to the constraints g(x,y)=0 and h(x,y)=0

    The constraints are supposed to be equations not inequalities

    So how can we apply Lagrange multipliers if the constraints are inequalities?
    Last edited by Idea; Nov 19th 2017 at 12:26 AM.
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    Re: Question about Lagrange multipliers for maximizing a function with two constraint

    Quote Originally Posted by Idea View Post
    The constraints are supposed to be equations not inequalities
    Yeah, so what’s happened there is that I’ve copied over the inequalities from a different method of optimization. But yes, the inequalities are supposed to be equalities for the sake of using Lagrange multipliers.
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    Re: Question about Lagrange multipliers for maximizing a function with two constraint

    Quote Originally Posted by johndoe69 View Post
    Yeah, so what’s happened there is that I’ve copied over the inequalities from a different method of optimization. But yes, the inequalities are supposed to be equalities for the sake of using Lagrange multipliers.
    so now the problem becomes trivial
    because the solution of the system of equations

    2x+y=44000

    y=20000

    is a single point (x=12000,y=20000)

    no need for Lagrange multipliers or anything else to find the max/min of f
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    Re: Question about Lagrange multipliers for maximizing a function with two constraint

    Quote Originally Posted by Idea View Post
    no need for Lagrange multipliers or anything else to find the max/min of f
    Could there be an excuse, however, to use Lagrange multipliers? I'm doing it as part of some school work, but I don't have time to rewrite the whole paper using different numbers and equations to make it not so trivial. I could just go ahead and explain after the fact that there was no need for Lagrange multipliers?
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