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Economics: find when utility maximized

  1. Jan 24, 2013 #1

    939

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    1. The problem statement, all variables and given/known data

    I must "maximize utility". This is done when the slope of -px/py = -MUx/MUy.

    MUx = (1/4x^-2/3)(y^2/5)
    MUy = (2/4x^1/3)(y^-1/5)
    Budget line: 80 = 7x + 2y

    2. Relevant equations

    MUx = (1/4x^-2/3)(y^2/5)
    MUy = (2/4x^1/3)(y^-1/5)
    Budget line: 80 = 7x + 2y

    3. The attempt at a solution
    -px/py = -7/2

    Thus, I must find when MUx/MUy = -7/2. I don't need the actual answer, just wondering how I would do this?
     
  2. jcsd
  3. Jan 24, 2013 #2

    Ray Vickson

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    Do you mean
    [tex] MU_x = \frac{1}{4} x^{-2/3} y^{2/5},[/tex]
    or do you mean
    [tex] MU_x = \frac{1}{4 x^{-2/3}} y^{2/5}?[/tex] If you mean the former, use parentheses, like this: MUx = (1/4) x^(-2/3) y^(2/5), but if you mean the latter, use parenthese also, but in different places, like this: MUx = 1/(4 x^(-2/3)) y^(2/5), etc.

    I will assume you want MUx = (1/4) x^(-2/3) y^(2/5) and MUy = (2/4) x^(1/3) y^(-1/5). So, just compute MUx/MUy using elementary algebra rules. Equating that to px/py gives one equation connecting x and y. The budget constraint gives another equation.

    If the expressions you give above are correct, the final set of 2 equations in 2 unknowns does not gave any simple "closed-form" solution; it must be tackled numerically.
     
  4. Jan 24, 2013 #3

    Mark44

    Staff: Mentor

    What do MUx and MUy mean? Each one is a function of both x and y, so what is the purpose of x in MUx and the purpose of y in MUy?

    I see where the -7/2 number comes from, which is solving the budget line equation for y (i.e., writing the line equation in slope-intercept form):
    7x + 2y = 80 ##\Rightarrow## y = (-7/2)x + 80.

    Is all that is asked for as simple as dividing MUx by MUy?
     
  5. Jan 24, 2013 #4

    939

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    Thanks, and I wanted the first one.

    They are different combinations of goods you can get within your budget constraint, i.e. in this case the budget constraint is 80, so you could buy so much of good x and so much of good y to remain in it, buy more of x, must buy less y, etc.

    ...

    I'm just wondering if it would be possible to solve this by mapping out all the budget constraints, and then substituting the x and y values in for MUx/MUy and see which one = -7/2?
     
  6. Jan 24, 2013 #5

    Ray Vickson

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    Yes, but that is doing it the hard way. However, go ahead and try it for yourself. I already suggested another method; use whichever one you like best.
     
  7. Jan 25, 2013 #6

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    I know, but I'm really bad at algebra :(... Can you give some advice here?

    7/2 = ((1/4) x^(-2/3) y^(2/5))/((2/4) x^(1/3) y^(-1/5))
    7/2 = 1/2x^-1 y^3/5
    ...
     
  8. Jan 25, 2013 #7

    Ray Vickson

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    So, you are saying
    [tex] \frac{7}{2} = \frac{1}{2} \frac{y^{3/5}}{x}.[/tex]
    What is stopping you from solving for x in terms of y? After doing that, you can plug in that expression for x into the budget constraint, and you will have a single equation containing y alone.
     
  9. Jan 25, 2013 #8

    939

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    Thanks, got it. I got it by solving x in terms of y.
     
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