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Thread: Lagrange Multipliers (Question)

  1. #1
    VkL
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    Lagrange Multipliers (Question)

    Hello, I am trying to find the absolute maximum and/or minimum of

    $\displaystyle f(x,y)=xy$ on the circle $\displaystyle x^2 + y^2=1$

    I am stuck and would like some help if any1 can help me.

    This is what I have so far,

    $\displaystyle \nabla f(x,y)=\lambda (g(x,y))$

    $\displaystyle \nabla f(x,y)=yi+xj$
    $\displaystyle \nabla g(x,y)=\lambda (2xi+2yj)$

    setting the components equal to each other:
    $\displaystyle y=2x\lambda$
    $\displaystyle x=2y\lambda$
    constraint: $\displaystyle x^2 + y^2=1$

    I am having trouble solving for $\displaystyle x, y, \lambda$

    I have 3 equations (including constraint) and 3 unknowns, It should work out. Can anyone help me? This is an algebra question, I feel like a moron.
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  2. #2
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    Hello, VkL!

    Sorry, I don't understand the method you were taught.


    Find extreme values of: .$\displaystyle f(x,y)\:=\:xy\:\text{ on the circle }\:x^2 + y^2\:=\:1$

    We have: .$\displaystyle F(x,y,\lambda) \;=\;xy + \lambda(x^2+y^2-1)$


    Set the partial derivative equal to zero and solve:

    . . $\displaystyle \begin{array}{ccccccc}F_x &=& y + 2x\lambda &=& 0 & [1] \\ \\[-3mm]
    F_y &=& x + 2y\lambda &=& 0 & [2] \\ \\[-3mm]
    F_{\lambda} &=& x^2+y^2-1 &=& 0 & [3] \end{array} $


    Solve [1] for $\displaystyle \lambda\!:\;\;\lambda \:=\:-\frac{y}{2x}\;\;[4]$

    Solve [2] for $\displaystyle \lambda\!:\;\;\lambda \:=\:-\frac{x}{2y}\;\;[5] $


    Equate [4] and [5]: .$\displaystyle -\frac{y}{2x} \:=\:-\frac{x}{2y} \quad\Rightarrow\quad y^2\:=\:x^2\quad\Rightarrow\quad y \:=\:\pm x\;\;[6]$


    Substitute into [3]: .$\displaystyle x^2+x^2-1 \:=\:0 \quad\Rightarrow\quad 2x^2\:=\:1 \quad\Rightarrow\quad x \:=\:\pm\frac{1}{\sqrt{2}} $

    Substitute into [6]: .$\displaystyle y \:=\:\pm\frac{1}{\sqrt{2}}$


    Therefore: .$\displaystyle \begin{Bmatrix}\text{Maximum:} & \left(\dfrac{1}{\sqrt{2}}\,,\:\dfrac{1}{\sqrt{2}}\ right) & \left(-\dfrac{1}{\sqrt{2}}\,,\:-\dfrac{1}{\sqrt{2}}\right) \\

    \\[-3mm] \text{Minimum:} & \left(\dfrac{1}{\sqrt{2}}\,,\;-\dfrac{1}{\sqrt{2}}\right) & \left(-\dfrac{1}{\sqrt{2}}\,,\:\dfrac{1}{\sqrt{2}}\right) \end{Bmatrix}$

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  3. #3
    MHF Contributor matheagle's Avatar
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    Both of these are the same.
    Creating this F is the same as making the two gradients a multiple of each other.
    The third equation is our constraint, g(x,y).
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  4. #4
    VkL
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    Soroban, Thank you very much for clearly explaining step by step! It was solving for lambda I got stuck on. I don't know why. Once again, thank you very much!
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  5. #5
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    Since the Lagrange multiplier method typically results in equations like $\displaystyle f_1(x,y,z)= \lambda g_1(x,y,z)$, $\displaystyle f_2(x,y,z)= \lambda g_2(x,y,z)$, etc. and the value of $\displaystyle \lambda$ itself is not relevant to the solution, I find it useful to divide one equation by another, immediately eliminating $\displaystyle \lambda$.

    In this case, setting $\displaystyle \nabla f(x,y,z)= \lambda\nabla g(x,y,z)$, which is the method I tend to use and is, of course, equivalent to Soroban's, we have $\displaystyle y= 2\lambda x$ and $\displaystyle x= 2\lambda y$.

    Dividing the first by the second, $\displaystyle \frac{y}{x}= \frac{2\lambda x}{2\lambda y}= \frac{x}{y}$ which immediately gives $\displaystyle x^2= y^2$ and so $\displaystyle x= \pm y$.

    Putting those into the constraint $\displaystyle x^2+ y^2= 1$ gives Soroban's solutions.
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