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Word problem using lagrange multiplier

  1. Jul 13, 2014 #1
    1. The problem statement, all variables and given/known data

    The Baraboo, Wisconsin plant of International Widget Co. uses aluminum, iron and magnesium to produce high-quality widgets. The quantity of widgets which may be produced using x tonnes of aluminum, y tonnes of iron and z tonnes of magnesium is Q(x,y,z) = xyz. the cost of raw materials is aluminum, $6 dollars per tonne; iron, $4 per tonne; magnesium, $8 per tonne. How many of each of aluminum, iron, and magnesium should be used to manufacture 1000 widgets at the lowest possible cost

    2. Relevant equations

    Q(x,y,z) = xyz is the objective function

    is the constraint 6x+4y+8z=1000 ?



    3. The attempt at a solution
     
    Last edited: Jul 13, 2014
  2. jcsd
  3. Jul 13, 2014 #2

    vela

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    No, it isn't. How did you come up with that? What is 6x+4y+8z supposed to represent? What does the 1000 represent? Does it make sense to set those two expressions equal to each other?
     
  4. Jul 13, 2014 #3
    I have issues with reading english.

    the 1000 represents the number of widgets

    6x,4y,8z would be the cost of each used material.

    now I see the issue because 1000 is not a cost, it is a quantity

    1000=xyz

    so I need to find my objective function

    would it not be the cost of each added up?
    6x+4y+8z=C(x,y,z) [doesn't make much sense for solving it either though when I try to actually try putting it together]
     
    Last edited: Jul 13, 2014
  5. Jul 13, 2014 #4

    vela

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    Yup, you're trying to minimize the cost subject to the constraint of a fixed quantity.
     
  6. Jul 13, 2014 #5

    Ray Vickson

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    Yes, your problem is to minimize 6x+4y+8z, subject to the constraint that xyz = 1000. Implicitly, you also need x,y,z > 0.

    The problem has a perfectly respectable solution that can be found via a Lagrange multiplier method, or by using the constraint to solve for z, say, in terms of x and y, then using that in the objective to get an unconstrained problem (although still with implicit constraints x,y > 0).
     
  7. Jul 13, 2014 #6
    thank you for the help, word problems are hard with my difficulty reading



    [tex]F(x,y,z,\lambda) = 6x+4y+8z - \lambda(xyz-1000)[/tex]

    [tex]\left\{\begin{array}{cc}6-yz\lambda=0 \\ 4-xz\lambda=0 \\8-xy\lambda=0 \\ -xyz+1000 =0 \end{array}\right.[/tex]

    [tex]\lambda = \frac{6}{yz} = \frac{4}{xz} = \frac{8}{xy} [/tex]

    whats the best step to take from here
     
    Last edited: Jul 13, 2014
  8. Jul 16, 2014 #7
    If I were able to isolate each equality for lambda to one variable then I could solve the restraint for lambda to get the x,y,z. and it seems if I try solving for any variable it will have two variable in the equality
     
  9. Jul 16, 2014 #8

    vela

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    What have you tried? It seems pretty straightforward how to solve this system.
     
  10. Jul 16, 2014 #9
    I tried solving for lambda first but that doesn't work.

    Then tried solving for z, but I didn't have luck. I'll try again this afternoon.
     
  11. Jul 16, 2014 #10

    vela

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    Show your work if you want help. Simply saying what you tried and then saying it didn't work is pretty much useless to us.
     
  12. Jul 16, 2014 #11
    I can't show it when I am replying in between classes. That why I said I will do it again this afternoon.
     
  13. Jul 16, 2014 #12
    [tex]F(x,y,z,\lambda) = 6x+4y+8z - \lambda(xyz-1000)[/tex]

    [tex]\left\{\begin{array}{cc}6-yz\lambda=0 \\ 4-xz\lambda=0 \\8-xy\lambda=0 \\ -xyz+1000 =0 \end{array}\right.[/tex]

    [tex]\lambda = \frac{6}{yz} = \frac{4}{xz} = \frac{8}{xy} [/tex]

    if I take

    [tex]6-yz\lambda=0; 4-xz\lambda=0[/tex]


    [tex]\lambda = \frac{6}{yz} = \frac{4}{zx}[/tex]

    solving for x [tex]x=\frac{2y}{3}[/tex] now using

    [tex]8-xy\lambda=0;4-xz\lambda=0 \Rightarrow \lambda = \frac{8}{xy}=\frac{4}{xz}[/tex]

    solving for z [tex]z=\frac{1y}{2}[/tex]

    now [tex] \frac{2y}{3}(y)\frac{1y}{2}=1000\Rightarrow \frac{1y}{3}=1000 \Rightarrow y =3000[/tex]

    [tex]x=2000;z=1500[/tex]
    this seems incorrect
     
  14. Jul 16, 2014 #13

    vela

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    ##\frac 23 y \times y \times \frac 12 y = \frac 13 y^3##, not ##\frac 13 y##.
     
  15. Jul 16, 2014 #14
    that's how tried I am. sorry

    so [tex]\frac{y^3}{3}=1000 \Rightarrow y=\sqrt[3]{1000/3}[/tex]
     
  16. Jul 16, 2014 #15

    vela

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    Not quite. Try again.
     
  17. Jul 16, 2014 #16
    I see whats wrong

    [tex] y= 10\sqrt[3]{3}[/tex]
     
    Last edited: Jul 16, 2014
  18. Jul 16, 2014 #17
    [tex] y= 10\sqrt[3]{3}; x = \frac{20\sqrt[3]{3}}{3};z=5\sqrt[3]{3}[/tex]
     
  19. Jul 16, 2014 #18

    vela

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    Looks good!
     
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