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Math Help - Lagrange Multipliers

  1. #1
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    Lagrange Multipliers

    The plane x+y+z=4 intersects the paraboloid z=x^2+y^2 in an ellipse. Find the points on the ellipse nearest to and farthest from the origin.

    So is the goal here to just minimize and then maximize the distance formula f(x,y,z)=\sqrt{x^2+y^2+z^2} with the constraints of g_1:x+y+z=4 and g_2:z=x^2+y^2?
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  2. #2
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    Hello, davesface!

    The plane x+y+z\:=\:4 intersects the paraboloid z\:=\:x^2+y^2 in an ellipse.
    Find the points on the ellipse nearest to and farthest from the origin.

    So is the goal to minimize and maximize the distance formula f(x,y,z)\:=\:\sqrt{x^2+y^2+z^2}
    with the constraints of g_1:x+y+z\:=\:4 and g_2:z\:=\:x^2+y^2 ? . . . . Yes!

    We have: . F(x,y,z,\lambda,\mu) \;=\;\left(x^2+y^2+z^2\right)^{\frac{1}{2}} + \lambda(x+y+z-4) + \mu\left(z-x^2-y^2\right)



    Find the five partial derivatives, equate to zero, and solve the system.

    . . \begin{array}{ccccc}\dfrac{\partial F}{\partial x} &=& \dfrac{x}{\sqrt{x^2+y^2+z^2}} + \lambda - 2\mu x &=& 0 \\ \\<br /> <br />
\dfrac{\partial F}{\partial y} &=& \dfrac{y}{\sqrt{x^2+y^2+z^2}} + \lambda - 2\mu y &=&0 \\ \\<br /> <br />
\dfrac{\partial F}{\partial z} &=& \dfrac{z}{\sqrt{x^21+y^2+z^2}} + \lambda + \mu &=&0 \end{array}

    . . . . \begin{array}{ccccc}\dfrac{\partial F}{\partial \lambda} &=& x + y + z - 4 &=& 0 \\ \\<br /> <br />
\dfrac{\partial F}{\partial \mu} &=& z - x^2 - y^2 &=& 0 \end{array}



    I'll wait in the car . . .
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  3. #3
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    I haven't ever seen someone take \frac{\delta F}{d \lambda} or \frac{\delta F}{d \mu} (and since those are constants, do those derivatives make sense?). According to my notes, it's set up as  \nabla f= \lambda \nabla g_1+ \mu \nabla g_2, which I think for this problem would be:

     \left [ \begin{array}{cc} \frac{x}{\sqrt{x^2+y^2+z^2}} \\ \frac{y}{\sqrt{x^2+y^2+z^2}}\\ \frac{z}{\sqrt{x^2+y^2+z^2}} \end{array} \right ]= \lambda   \left [ \begin{array}{cc} 1 \\ 1\\1 \end{array} \right ] + \mu  \left [ \begin{array}{cc} 2x \\ 2y \\-1 \end{array} \right ]
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