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Math Help - linear programming question

  1. #1
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    linear programming question

    A fruit drink is made from 3 ingredients.

    let j be the fraction of the drink that is juice
    s be the fraction of the drink that is sugar and flavouring
    w be the fraction of the drink that is water.

    the cost of the drink per liter is $12j+3s+w, and it can sold at $2+s+j/litre. choose j and s to maximize the profit, for marketing reasons j>=1/20

    1. express profit per litre as a function of j and s.
    How do express profit, say i let z=profit in dollars per litre. z= price sold-cost=2-11j-2s????

    2. Express problem to maximize profit, subject to constraints.
    well my constraints are j>=1/20, s,j,w>=0... what are my others?
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  2. #2
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    Quote Originally Posted by fredrick08 View Post
    A fruit drink is made from 3 ingredients.

    let j be the fraction of the drink that is juice
    s be the fraction of the drink that is sugar and flavouring
    w be the fraction of the drink that is water.

    the cost of the drink per liter is $12j+3s+w, and it can sold at $2+s+j/litre. choose j and s to maximize the profit, for marketing reasons j>=1/20

    1. express profit per litre as a function of j and s.
    How do express profit, say i let z=profit in dollars per litre. z= price sold-cost=2-11j-2s????

    2. Express problem to maximize profit, subject to constraints.
    well my constraints are j>=1/20, s,j,w>=0... what are my others?
    1. \text{Profit per litre} = \text{revenue - cost} =  (2+j+s) - (12j+3s+w)=2-11j-2s-w

    2. As s,j,w are fractions by volume you also have s+j+w=1
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  3. #3
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    ok i drew it out, and done it in solver, but its says that there is no water?? Is this true? and my optimal solution is at (0.95,0,05) profit=1.55??
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  4. #4
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    i did the math and subbed j=1/20 and s=19/20, then profit=-9/20??? im really confused now, how did solver get 1.55??
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by fredrick08 View Post
    ok i drew it out, and done it in solver, but its says that there is no water?? Is this true? and my optimal solution is at (0.95,0,05) profit=1.55??
    You should be maximising profits (looks like you have minimised?)

    By inspection it looks like maximum profit is returned with j=0.05, w=0.95, s=0. When the profit is $0.5

    CB
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  6. #6
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    yes, that is what I thought, but my lecturer told me to use j+s+w=1, to eliminate w.... i got no idea what that means. im jet gunna leave it is profit=0.5
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  7. #7
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    Quote Originally Posted by fredrick08 View Post
    yes, that is what I thought, but my lecturer told me to use j+s+w=1, to eliminate w.... i got no idea what that means. im jet gunna leave it is profit=0.5
    In the profit expression replace w by 1-j-s

    CB
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  8. #8
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    i see, but how am i meant to know when i draw the feasible region that s=0?
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  9. #9
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    Quote Originally Posted by fredrick08 View Post
    i see, but how am i meant to know when i draw the feasible region that s=0?
    Draw the feasible region, it is defined by the constraints s\ge 0, j\ge 0.05, s+j\le 1.

    Now the optimum (maximum profit) occurs at a vertex of the feasible region, so check them and keep the one with maximum profit.

    CB
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