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Math Help - lagrange multiplier

  1. #1
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    lagrange multiplier

    I need to know if I'm solving for Lagrange multiplier correctly.

    f(x,y,z)= 8x-4y; \ x^2+10y^2+z^2=5

    so far I have:

    \begin{array}{l}f_x = 8, \ f_y = -4 \ f_z = 0 \\ \\ <br />
f_x = 2x, \ f_y = 20y \ f_z = 2z \\ \\<br />
8=\lambda 2x \rightarrow \lambda = \frac{4}{x} \\ \\<br />
-4 = \lambda 2y \rightarrow -1 = \lambda 5y = \left( \frac{4}{x} \right) 5y \ \rightarrow y = -\frac{x}{20}\\ \\<br />
0 = \lambda z \rightarrow z= 0<br />
\end{array}<br />

    putting the above values in the constraint I get:

    x^2+10 \left( -\frac{x}{20} \right) ^2+(0)^2=5 = \frac{41}{40} x^2 \rightarrow x = \pm \sqrt{ \frac{200}{41}}

    then just substitute for y...

    is this correct?
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  2. #2
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    Quote Originally Posted by lllll View Post
    I need to know if I'm solving for Lagrange multiplier correctly.

    f(x,y,z)= 8x-4y; \ x^2+10y^2+z^2=5

    so far I have:

    \begin{array}{l}f_x = 8, \ f_y = -4 \ f_z = 0 \\ \\ <br />
f_x = 2x, \ f_y = 20y \ f_z = 2z \\ \\<br />
8=\lambda 2x \rightarrow \lambda = \frac{4}{x} \\ \\<br />
-4 = \lambda 2y \rightarrow -1 = \lambda 5y = \left( \frac{4}{x} \right) 5y \ \rightarrow y = -\frac{x}{20}\\ \\<br />
0 = \lambda z \rightarrow z= 0<br />
\end{array}<br />


    putting the above values in the constraint I get:

    x^2+10 \left( -\frac{x}{20} \right) ^2+(0)^2=5 = \frac{41}{40} x^2 \rightarrow x = \pm \sqrt{ \frac{200}{41}}

    then just substitute for y...

    is this correct?
    I did it a little differently and came up with

    x=\frac{10\sqrt{82}}{41}

    It is the same as yours

    Nice work.!!
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