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Math Help - Algebra Maximizing Function

  1. #1
    neoseeker191
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    Algebra Maximizing Function

    Maximize the function P=2x+3y

    with the following constraints:

    X is greater than or equal to 0
    0 is less than or equal to y is less than or equal to 1
    3x+y is less than or equal to 3
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  2. #2
    Super Member wingless's Avatar
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    Let's write the conditions.
    x \geq 0
    0 \leq y \leq 1
    3x + y \leq  3
    And we want to maximize: P = 2x + 3y

    We have to maximize both x and y to get a maximum 2x + 3y.
    As 3x + y \leq  3 and we want to maximize x and y, making it less than 3 won't make x and y maximum. So, 3x + y = 3

    x can be any positive integer but y is between (or equal to) 0 and 1. This means finding y is better cuz when we use y, we don't have to think of x, as it can be any positive integer.

    3x + y = 3
    x = \frac {3-y}{3}

    Then, P = 2x + 3y
    P = 2\frac {3-y}{3} + 3y
    P = 2 + \frac {7y}{3}

    The greater y will make our P bigger. But we have a condition for y, 0 \leq y \leq 1. The greatest y value we can use is 1. For y=1, x is 2/3. So P is \frac {13}{3}.
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  3. #3
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    Hello, neoseeker191!

    Maximize the function: . P\:=\:2x+3y
    with the following constraints: . x \:\geq 0,\quad 0 \leq y \leq 1, \quad 3x+y \:\leq \:3
    This is a Linear Programming problem.
    You should have been shown the approach by now . . .

    Graph the inequalities; shade the appropriate regions.
    Determine the vertices of the resulting polygon.
    Test the vertices (coordinates) in the given function.

    The graph should look like this:
    Code:
            |
          3 |
            |\
            | \
            |  \
            |   \
            |    \
          1 o - - o - -
            |::::::\
            |:::::::\
            |::::::::\
          - o - - - - o - -
            |         1\

    Determine the four vertices.
    You should get: . (0,\,0)\,;(1,\,0),\;(0,\,1),\;\left(\frac{2}{3},\,  1\right)

    Now see which one produces maximum P.

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