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Math Help - Help with a LaGrange Multiplier Question

  1. #1
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    Help with a LaGrange Multiplier Question

    Hi. I am trying to maximize the function F(x,y,z)=xyz, under the constraint 2x +2y+z=84, with theta being represented by m.

    Derivatives:

    Fx= yz+2m=0
    Fy= xz+2m=0
    Fz= xy+m=0
    Fm= 2x+2y+z-84=0.

    I found m= -yz/2, and m= -xz/2, getting y=x. I proceeded to find z= -2y.

    I then plugged everything into Fm and got y=42. What am I doing wrong, as the correct answer should be (14,14,28)?

    Many thanks
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  2. #2
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    Hello, jaijay32!

    You were close . . . one little error.


    Maximize the function F(x,y,z)\,=\,xyz
    under the constraint 2x +2y+z\:=\:84


    Derivatives:

    . . \begin{array}{cccccc}(1) & F_x &=& yz+2\theta &=& 0 \\<br />
(2) & F_y &=& xz+2\theta &=& 0 \\<br />
(3) & F_z &=&  xy+\theta &=& 0 \end{array} . . Correct!
    . . (4)\;\;\;F_{\theta} \;\;=\;\; 2x+2y+z-84 \;=\; 0


    I found: . (5)\;\theta\,=\, -\frac{yz}{2}\,\text{ and }\,(6)\;\theta \,=\, -\frac{xz}{2}, getting y\,=\,x . Right!

    I proceeded to find z\,=\, -2y .This is wrong.

    From (3), we have: . \theta \:=\:-xy

    Equate to (6): . -xy \:=\:-\frac{xz}{2} \quad\Rightarrow\quad z \:=\:2y

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