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Math Help - Lagrange multipliers

  1. #1
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    Lagrange multipliers

    Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

    f(x,y) = 4x + 6y; x^2 + y^2 = 13

    I'm not sure how to solve this.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by paulrb View Post
    Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

    f(x,y) = 4x + 6y; x^2 + y^2 = 13

    I'm not sure how to solve this.
    Do you know how to use Lagrange Multipliers? Basically let us define f(x,y)=4x+6y and g(x,y)=x^2+y^2-13. We then need to solve the system of equations given by

    \left\{\begin{array}{cc}f(x,y)_x=\lambda{g(x,y)_x}  \\f(x,y)_y=\lambda{g(x,y)_y}\\g(x,y)=0\end{array}\  right\}

    A little work shows the above is equivalent to

    \left\{\begin{array}{cc}4=2x\lambda\\6=2y\lambda\\  x^2+y^2+-13=0\end{array}\right\}

    And unless my calculations are off I calculate the solutions to be

    \left\{\begin{array}{cc}\left\{x,y,\lambda\right\}  =\left\{-2,-3-1\right\}\\\left\{x,y,\lambda\right\}=\left\{2,3,1  \right\}\end{array}\right\}

    So now f(-2,-3)=-26 and f(2,3)=26. From there we can conclude that the point (-2,-3,-26) is a minimum point and (2,3,26) a maximum point.
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  3. #3
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    Hello, paulrb!

    I must assume you know the basic steps for Lagrange multipliers.


    Use Lagrange multipliers to find the maximum and minimum values
    of the function subject to the given constraints.

    . . f(x,y) \:= \:4x + 6y \qquad x^2 + y^2 \:=\: 13

    We have: . F(x,y,\lambda) \;=\;4x + 6y + \lambda(x^2+y^2-13)


    Take partial derivatives and equate to zero.

    . . F_x \;=\;4 + 2x\lambda\:=\:0 \quad\Rightarrow\quad x \:=\:-\frac{2}{\lambda} .[1]

    . . F_y \;=\;6 + 2y\lambda \:=\:0 \quad\Rightarrow\quad y \:=\:-\frac{3}{\lambda} .[2]

    . . F_{\lambda} \;=\;x^2-y^2-13 \:=\:0 \quad\Rightarrow\quad x^2+y^2 \:=\:13 .[3]


    Substitute [1] and [2] into [3]: . \left(-\frac{2}{\lambda}\right)^2 + \left(-\frac{3}{\lambda}\right)^2 \:=\:13 \quad\Rightarrow\quad \frac{4}{\lambda^2} + \frac{9}{\lambda^2} \:=\:13<br />

    . . . . . . \frac{13}{\lambda^2} \:=\:13 \quad\Rightarrow\quad \lambda^2 \:=\:1\quad\Rightarrow\quad\lambda \:=\:\pm1

    Substitute into [1]: . x \:=\:-\frac{2}{\pm1} \quad\Rightarrow\quad x \:=\:\pm2

    Substitute into [2]: . y \:=\:-\frac{3}{\pm1} \quad\Rightarrow\quad y \:=\:\pm3


    . . \begin{array}{cccccc}f(2,3) &=& 4(2) + 6(3) &=& 26 & \text{maximum} \\ \\[-3mm]<br />
f(\text{-}2,\text{-}3) &=& 4(\text{-}2) + 6(\text{-}3) &=& \text{-}26 & \text{minimum} \end{array}

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