How can the owners maximize profits by pricing their coffee beans effectively?

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Homework Statement


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?


Homework 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.


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.
 
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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.
 
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?
 
jheld said:

Homework Statement


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?


Homework 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.


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.
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 P_M= (x- 2)(80- 100x- 40x). 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 P_K= (y-4)(20+ 6x- 35y). Those are the functions you need to differentiate, giving two equations to solve for x and y.
 
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.
 
It's on the right track, but to finish it off, you should also apply the second derivative test for partial deriviatives.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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