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My maths, they need solution'd again!

TL DRTL DR Not at all confident in his reflexive opinions of thingsRegistered User regular
edited November 2007 in Help / Advice Forum
You guys helped me ace my math test last time around, now I must call upon you again.

I'm constructing a table to algebraically solve this problem:
Maximize P = 3x + 4y + 2z
Subject to x + y + z <= 10
x + 2y +2z <= 14

I rewrite with slack variables S and T, and all is going well, until I have to find the values for Y and Z when X, S, and T are zero variables.

So my problem would look like this:
y + z = 10
2y + 2z = 14

And I am stuck.


All help is appreciated, thanks in advance.

TL DR on

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    AphostileAphostile San Francisco, CARegistered User regular
    edited November 2007
    Solve for y or z in one equation, substitute it into the second equation, solve that one for a single variable, use that single variable in the first equation to solve for the other variable.

    So, y = 10 - z
    Plug in 2(10-z) + 2 z = 14
    Solve accordingly?

    Edit: I realized that my math yields a zero. I think you must've mixed something up earlier on, because y + z = 10 and y+z = 7

    Aphostile on
    Nothing. Matters.
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    TL DRTL DR Not at all confident in his reflexive opinions of thingsRegistered User regular
    edited November 2007
    Well the full equations would be

    x + y + z + s = 10
    x + 2y+ 2z+ t = 14

    And when x, s, and t all = 0, we're left with

    0 + y + z + 0 = 10
    0 + 2y+ 2z+ 0 = 14

    So taking you're suggestion and discarding the zero variables, I'd get something like

    y = 10 - z

    which, plugged into the 2nd equation, results in

    2 (10 - z) +2z = 14
    or
    20 = 14

    O_o:|

    TL DR on
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    TL DRTL DR Not at all confident in his reflexive opinions of thingsRegistered User regular
    edited November 2007
    I'll come back to that problem in a minute.

    This one is asking me to minimize c = 16x + 42y + 5z
    subject to 2x + 6y >= 20
    2x + 3y + z >= 40
    by finding the dual.

    Dual:
    Maximize P= 20x + 40x
    Subject to 2x + 2y <= 16
    6x + 3y <= 42
    y <= 5
    Pivot, pivot, and I get
    x = 9/2
    y = 5
    p = 290


    Not hard. I just can't figure out the deal with that damn table in the other problem.

    TL DR on
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    enlightenedbumenlightenedbum Registered User regular
    edited November 2007
    This is sort of muddled, but I'm reasonably sure it works:

    x + y + z <= 10
    x + 2y + 2z <= 14

    Make the assumption that the solution will come from the boundaries (it will, cause there's only one critical point and yay calculus):

    x + y + z = 10
    x + 2y + 2z = 14

    Or:

    2x + 2y + 2z = 20
    x + 2y + 2z = 14

    Which leads to:

    x = 6

    So then y + z = 4 from either of our equations above.

    Or: y = 4 - z

    So: P = 18 + 16- 4z + 2z
    = 34 - 2z

    Making the assumption that the variables cannot be nonnegative, z then = 0, y = 4 and P's maximum value is 34.

    Alternately:

    It's clear that x <= 6 and y + z <= 4, right? Because otherwise these equations wouldn't be used to their fullest potential and that's no good at all.

    So you've got: P = 3x + 4y + 2z

    Group it instead as follows:

    P = 3x + 2y + 2(y + z)

    The first term is maximized if x = 6, the third term is maximized when y + z = 4, and the second term is maximized when y is the biggest it can be such that y + z <= 4, i.e. y = 4.

    And again we get x = 6, y =4, z = 0 and P = 34 as the maximum. However, again we assume the answer only has non-negative values for x, y, and z.

    Hope that helps.

    enlightenedbum on
    Self-righteousness is incompatible with coalition building.
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    Blake TBlake T Do you have enemies then? Good. That means you’ve stood up for something, sometime in your life.Registered User regular
    edited November 2007
    This is sort of muddled, but I'm reasonably sure it works:

    x + y + z <= 10
    x + 2y + 2z <= 14

    Make the assumption that the solution will come from the boundaries (it will, cause there's only one critical point and yay calculus):

    x + y + z = 10
    x + 2y + 2z = 14

    Or:

    2x + 2y + 2z = 20
    x + 2y + 2z = 14

    Which leads to:

    x = 6

    So then y + z = 4 from either of our equations above.

    Or: y = 4 - z

    So: P = 18 + 16- 4z + 2z
    = 34 - 2z

    Making the assumption that the variables cannot be nonnegative, z then = 0, y = 4 and P's maximum value is 34.

    Alternately:

    It's clear that x <= 6 and y + z <= 4, right? Because otherwise these equations wouldn't be used to their fullest potential and that's no good at all.

    So you've got: P = 3x + 4y + 2z

    Group it instead as follows:

    P = 3x + 2y + 2(y + z)

    The first term is maximized if x = 6, the third term is maximized when y + z = 4, and the second term is maximized when y is the biggest it can be such that y + z <= 4, i.e. y = 4.

    And again we get x = 6, y =4, z = 0 and P = 34 as the maximum. However, again we assume the answer only has non-negative values for x, y, and z.

    Hope that helps.

    Acutally you are correct and it's not an assumption since the question asks you to maximise P and if you use negative values it will bring down the value of P making the solution invalid.

    Blake T on
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    enlightenedbumenlightenedbum Registered User regular
    edited November 2007
    Blaket wrote: »
    This is sort of muddled, but I'm reasonably sure it works:

    x + y + z <= 10
    x + 2y + 2z <= 14

    Make the assumption that the solution will come from the boundaries (it will, cause there's only one critical point and yay calculus):

    x + y + z = 10
    x + 2y + 2z = 14

    Or:

    2x + 2y + 2z = 20
    x + 2y + 2z = 14

    Which leads to:

    x = 6

    So then y + z = 4 from either of our equations above.

    Or: y = 4 - z

    So: P = 18 + 16- 4z + 2z
    = 34 - 2z

    Making the assumption that the variables cannot be nonnegative, z then = 0, y = 4 and P's maximum value is 34.

    Alternately:

    It's clear that x <= 6 and y + z <= 4, right? Because otherwise these equations wouldn't be used to their fullest potential and that's no good at all.

    So you've got: P = 3x + 4y + 2z

    Group it instead as follows:

    P = 3x + 2y + 2(y + z)

    The first term is maximized if x = 6, the third term is maximized when y + z = 4, and the second term is maximized when y is the biggest it can be such that y + z <= 4, i.e. y = 4.

    And again we get x = 6, y =4, z = 0 and P = 34 as the maximum. However, again we assume the answer only has non-negative values for x, y, and z.

    Hope that helps.

    Acutally you are correct and it's not an assumption since the question asks you to maximise P and if you use negative values it will bring down the value of P making the solution invalid.

    It totally wouldn't! Check out: y = 100, z = - 96 and x = 6. It still fits all the constraints, but suddenly P is huge! namely, 18 + 400 - 192 = 226.

    enlightenedbum on
    Self-righteousness is incompatible with coalition building.
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    Blake TBlake T Do you have enemies then? Good. That means you’ve stood up for something, sometime in your life.Registered User regular
    edited November 2007
    Oh shit sorry my bad. For some reason I just glanced at P and thought it was the sumation not an actual formula.

    Blake T on
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    FunkyWaltDoggFunkyWaltDogg Columbia, SCRegistered User regular
    edited November 2007
    I am a little rusty, but I believe the y + z = 10, 2y + 2z = 14 problem arises because the equation has no maximum under those constraints. It's easy to see, as enlightenedbum pointed out: pick y as large as you like, and let z = -y. z cancels y in both the constraints, but 4y overpowers 2z = -2y.

    FunkyWaltDogg on
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