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Maximizing Coffee Profits

  1. Jun 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Mocha beans are priced at X dollars per pound, and Kona beans at y dollars per pound.
    80 - 100x + 40y pounds of Mocha beans sold each week
    20 + 60x -35y Kona beans sold each week. Cost of the beans is $2 per lbs of Mocha and $4 of Kona beans to the owners.
    How should the owners price the coffee beans in order to maximize their profits?

    2. Relevant equations
    Looking for critical points (maximums, extremum).
    Take the first derivative of each partial, see what the variable should be for it to be equal to 0.
    Then go to the second partials and put them in the Hessian matrix, and solve with the derivative to see what the max should be.

    3. The attempt at a solution
    Solving for Mocha first..
    f_x = -100
    f_y = 40
    But after that, I get stuck; I am not sure how to relate the $2 cost we were given.
  2. jcsd
  3. Jun 30, 2009 #2


    Staff: Mentor

    What do x and y represent? Your first sentence indicates that these are the per-pound prices, respectively, of Mocha and Kona coffee beans, but then you have two expressions that also involve x and y for the pounds of each type sold. I don't see how the variables in your two weight expressions can be the same as those of the per-pound price.
  4. Jul 5, 2009 #3
    Yeah, it is phrased weird. I'll rewrite the entire problem statement just to be sure:
    The Java Joint Gourmet Coffee House sells top-of-the-line Arabian Mocha and Hawaiian Kona beans. If Mocha beans are priced at x dollars per pound and Kona beans at y dollars per pound, then market research has shown that each week approximately 80 - 100x + 40y pounds of Mocha benas will be sold and 20 + 60x - 35y pounds of Kona beans will be sold. The wholesale cost to the Java Joint owners is $2 per pound for Mocha beans and $4 per pound for Kona beans. How should the owners price the coffee beans in order to maximize their profits?
  5. Jul 5, 2009 #4


    User Avatar
    Science Advisor

    How did you get "f_x= -100"? There is no "f" in the problem! Don't just start taking derivatives of what ever formula you see.

    The owners want to maximize profits. If x is the price of a pound of Mocha beans and each pound cost $2, then they make a profit of x- 2 dollars on each pound of Mocha beans. If they sell 80 - 100x + 40y pounds of Mocha beans, then the total profit from Mocha beans is "profit per pound times pounds" or [itex]P_M= (x- 2)(80- 100x- 40x)[/itex]. If Y is the price of a pound of Kona beans and each pound cost $4, they make a profit of y- 4 dolars on each pound of Kona beans. If they sell 20 + 60x -35y pounds of Kona beans, they make a profit of [itex]P_K= (y-4)(20+ 6x- 35y)[/itex]. Those are the functions you need to differentiate, giving two equations to solve for x and y.
  6. Jul 5, 2009 #5
    Okay, so far I understand. And, just to clarify, I wasn't using f_x and f_y notation for the hell of it, I was just making f the general equation. But, I understand the x - 2 and y -4 in conjunction with the two equations given.

    So, I have taken the first (and second order) derivatives of each equation, but from there I am a little stuck. I have a few different points, and thus I am unsure of where they connect.

    P_M(x) = -120 - 200x + 40y, where x = 0, y = 3;
    P_M(y) = 40x - 80, where x = 2, y = 0;

    P_K(x) = 60y - 240, where y = 4;
    P_K(y) = 60x - 70y + 160, where x = -8/3, y = 16/7

    I have also calculated the 2nd order partial derivatives, but I wanted to make sure that this was on the right track.
  7. Jul 6, 2009 #6
    It's on the right track, but to finish it off, you should also apply the second derivative test for partial deriviatives.
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