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Math Help - Least cost calculation, plz help!

  1. #1
    Remeth
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    Red face Least cost calculation, plz help!

    Need immediate help on this question. Any help is appreciated!


    Cost formulas for objects A and B

    A) 3A^2 + 100A+400
    B) 6B^2+150B+600
    A+B = 500

    ^2 means to square

    What is the combination of A and B that would yield the lowest cost? I don't need the answer. Just the formula if there is one or the steps to take to get to the answer. Thanks a lot
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Remeth View Post
    Need immediate help on this question. Any help is appreciated!


    Cost formulas for objects A and B

    A) 3A^2 + 100A+400
    B) 6B^2+150B+600
    A+B = 500

    ^2 means to square

    What is the combination of A and B that would yield the lowest cost? I don't need the answer. Just the formula if there is one or the steps to take to get to the answer. Thanks a lot
    Total cost:

    T = 3A^2 + 100A+400 + 6B^2+150B+600

    but B=500-A, so substituting this in the formula for T we have:

    T = 9 A^2 - 6050 A + 1576000.

    To find the lowest cost we differentiate T with respect to A and set the
    derivative equal to zero and solve for A.

    d/dA T = 6 A - 6050,

    so for the minimum: A=6050/6, and B=500 - 6050/6. This gives the point
    of global minimum for T (since T is a quadratic in A opening upwards this
    stationary point is of necessity an global minimum rather than any other
    king of stationary point).

    Now I assume that we have to have A,B >=0, so this minimum is not
    in the feasible region for this problem, so with these additional constraints
    we know that the minimum lies on the boundard of the feasible region.

    In this case the feasible region is the line segment from (0,500) to (500,0)
    in the AB plane, so the boundary of the feasible region is the pair of points
    (500,0) anf (0,500).

    That is either A=0 or B=0, When B=0 T=801000, when A=0 T =2026000, so
    the minimum occurs when A=500 and B=0, and the total cost is 801000.

    RonL
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Total cost:

    T = 3A^2 + 100A+400 + 6B^2+150B+600

    but B=500-A, so substituting this in the formula for T we have:

    T = 9 A^2 - 6050 A + 1576000.

    To find the lowest cost we differentiate T with respect to A and set the
    derivative equal to zero and solve for A.

    d/dA T = 6 A - 6050,

    so for the minimum: A=6050/6, and B=500 - 6050/6. This gives the point
    of global minimum for T (since T is a quadratic in A opening upwards this
    stationary point is of necessity an global minimum rather than any other
    king of stationary point).

    Now I assume that we have to have A,B >=0, so this minimum is not
    in the feasible region for this problem, so with these additional constraints
    we know that the minimum lies on the boundard of the feasible region.

    In this case the feasible region is the line segment from (0,500) to (500,0)
    in the AB plane, so the boundary of the feasible region is the pair of points
    (500,0) anf (0,500).

    That is either A=0 or B=0, When B=0 T=801000, when A=0 T =2026000, so
    the minimum occurs when A=500 and B=0, and the total cost is 801000.

    RonL
    I was thinking of doing something along these same lines, but I wasn't sure if he knows calculus. Doing this the algibraic way would involve completing the square, which would be nasty in this case.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ecMathGeek View Post
    I was thinking of doing something along these same lines, but I wasn't sure if he knows calculus. Doing this the algibraic way would involve completing the square, which would be nasty in this case.
    It had crossed my mind that maybe a non calculus minimisation of the
    quadratic might be appropriate, but then if the additional constraint that
    A and B>=0 is correct, then this is unlikely to be a precalculus problem.

    But meybe I'm wrong!

    RonL
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