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Math Help - Maximising profit

  1. #1
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    Maximising profit

    A company operates two cafes. The total daily cost of running the two cafes is
    TC = 6000 + 180x + 140y
    where x is the number of patrons in cafe one and y the number in cafe two. The total daily revenue from the two cafes is
    R = 320x + 250y - 0.375x2 - 0. 5y2 - 0.25xy
    Determine the level of patronage required to maximise total profit. How much are these profits?
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  2. #2
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    Quote Originally Posted by brumby_3 View Post
    A company operates two cafes. The total daily cost of running the two cafes is
    TC = 6000 + 180x + 140y
    where x is the number of patrons in cafe one and y the number in cafe two. The total daily revenue from the two cafes is
    R = 320x + 250y - 0.375x2 - 0. 5y2 - 0.25xy
    Determine the level of patronage required to maximise total profit. How much are these profits?
    Profit:

    P(x,y)=R-TC=140x+110y-0.375x^2-0.5y^2-0.25 xy-6000<br />

    Now you need to find the critical point/s of P(x,y) (that is points where both partial derivatives vanish) and identify which of these gives a maximum (if there is more than one, if there is one confirm that the profit is a maximum there).

    RonL
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    Can someone further step-by-step show me how to solve this problem so I can try for some other questions I have please?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    Profit:

    P(x,y)=R-TC=140x+110y-0.375x^2-0.5y^2-0.25 xy-6000<br />

    Now you need to find the critical point/s of P(x,y) (that is points where both partial derivatives vanish) and identify which of these gives a maximum (if there is more than one, if there is one confirm that the profit is a maximum there).

    RonL
    P(x,y) is a quadratic surface and so there should be exactly one critical point, and it is either a maximum or a minimum for P. Ramining the form of P shows that in this case it is a maximum.

    \frac{\partial}{\partial x}P(x,y)= 140-0.75x-0.25y

    \frac{\partial}{\partial y}P(x,y)= 110-y-0.25x

    So the critical point is the solution of the simultaneous equations:

    0.75x+0.25y=140

    0.25x +y = 110

    There is probably a non-calculus method for this problem which involves completing the square for the profit, but I would have to think a bit about how that would work.

    RonL
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