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Math Help - Domain in optimization problem

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    Domain in optimization problem

    #11. There are $320 available to fence in rectangular garden. The fencing for the side of the garden facing the road costs $6 per foot and the fencing for the other three sides costs $2 per foot. Consider the problem of finding the dimensions of the largest possible garden.

    So I solved everything and got:
    A=xy as the objective equation
    C=8y+4x=320 as the constraint equation

    A=(80x-x^2)/2
    A'=(80-2x)/2 then I set that to zero and got x=40, y=20.

    How come the correct answer is x=20 y=40? More importantly, how do you get the domain? (which is 0<x<40)

    * i realized that I set C=8y+4x=320 while the answer key has C=8x+4y=320* Would my answer be still correct?
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    Quote Originally Posted by fabxx View Post
    #11. There are $320 available to fence in rectangular garden. The fencing for the side of the garden facing the road costs $6 per foot and the fencing for the other three sides costs $2 per foot. Consider the problem of finding the dimensions of the largest possible garden.

    So I solved everything and got:
    A=xy as the objective equation
    C=8y+4x=320 as the constraint equation

    A=(80x-x^2)/2
    A'=(80-2x)/2 then I set that to zero and got x=40, y=20.

    How come the correct answer is x=20 y=40? More importantly, how do you get the domain? (which is 0<x<40)

    * i realized that I set C=8y+4x=320 while the answer key has C=8x+4y=320* Would my answer be still correct?
    It's just a matter of whether x has to be the shorter side or not. You will agree that a 20 by 40 rectangle is the same as a 40 by 20 rectangle. So your answer is correct.... *unless* the problem specifically tells you x must be the shorter side.
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    Quote Originally Posted by fabxx View Post
    More importantly, how do you get the domain? (which is 0<x<40)
    Sorry, forgot this part.

    The domain follows easily from C=8y+4x=320, or C=8x+4y=320, however you choose to write it.

    I'll choose C=8x+4y=320 just to avoid confusion. Say x = 40. Then

    8*40+4y=320

    320+4y = 320

    4y = 0

    y = 0

    See the problem?
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    i still don't get how you got the domain. so if the equation is 8y+4x=320 and I plug in x=40, that makes y=20 (which is exactly what y is)

    but how do you figure out that the domain is 0<x<40?
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    Quote Originally Posted by fabxx View Post
    i still don't get how you got the domain. so if the equation is 8y+4x=320 and I plug in x=40, that makes y=20 (which is exactly what y is)

    but how do you figure out that the domain is 0<x<40?
    If you write 8y+4x=320 then you will need to plug in y=40. In this case, we have the restriction 0 < y < 40. All you did was switch the variable names.

    The letter "x" isn't special, it's just a convention to name an independent variable x and a dependent variable y.

    So we can validly write x = f(y) = (320-8y)/4 and ask for the domain (keeping in mind there is a requirement that x > 0), then find that the domain is 0 < y < 40.

    Make sense?
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    Okay, you haven't responded yet, but I'm worried that maybe you're still confused, so maybe this will help.

    We can look at this problem a few different ways. For completeness, I'll just type out all 4 ways so hopefully you see how it all fits.

    Situation 1: We chose to write 8x+4y=320, and we let y = f(x).

    This is the book answer.

    In this case, the domain is 0 < x < 40 and the range is 0 < y < 80.

    Situation 2: We chose to write 8x+4y=320, and we let x = f(y).

    Then the domain is 0 < y < 80 and the range is 0 < x < 40.

    Situation 3: We chose to write 8y+4x=320, and we let y = f(x).

    Then the domain is 0 < x < 80 and the range is 0 < y < 40.

    Situation 4: We chose to write 8y+4x=320, and we let x = f(y).

    Then the domain is 0 < y < 40 and the range is 0 < x < 80.

    See how it's always the same, but with certain things switched with each other.

    I hope it's clear now. If not, just ask.
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