MATH

1. A factory manufactures two products, each requiring the use of three machines. The first machine can be used at most 60 hours; the second machine at most 30 hours; and the third machine at most 80 hours. The first product requires 2 hours on machine 1, 1 hour on machine 2, and 1 hour on machine 3; the second product requires 1 hours each on machine 1 and 2, and 3 hours on machine 3. If the profit is RM30 per unit for the first product and RM50 per unit for the second product, how many units of each product should be manufactured to maximize profit?

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  1. let the number of the first product be x
    let the number of the 2nd product by y

    So we need to look at the regions
    2x+y ≤60 and
    x+y≤30 and
    x+3y≤80

    The profit function is 30x+50y
    which has a slope of -3/5

    move that profit line to the right, parallel to itself, until you reach the farthest vertex of that region.
    From my rough sketch that appears to be the
    intersection of
    2x+y = 60 , ---> y = 60-2x
    and
    x + 3y = 80

    substitution:
    x + 3(60-2x) = 80
    x-6x = -100
    x = 20
    then y = 60-20 = 40

    max profit is 30(20) + 50(40) or 2600 , when 20 of the first product and 40 of the second product are made

    Check: intersect the other two lines for the other vertex
    x+y=30
    x+3y=80
    subtract them
    2y = 50
    y = 25 , then x = 5
    profit = 30(5) + 50(25) = only 1400

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    posted by Reiny
  2. just noticed an arithmetic error..

    in my evaluation of y, should be
    60 - 2x
    = 60-2(20)
    = 20

    so .....

    max profit is 30(20) + 50(20) or 1600 , when 20 of the first product and 20 of the second product are made

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    posted by Reiny

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