Thread: 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?

Originally Posted by brumby_3

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:



Now you need to find the critical point/s of (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

Can someone further step-by-step show me how to solve this problem so I can try for some other questions I have please?

Originally Posted by CaptainBlack

Profit:



Now you need to find the critical point/s of (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

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





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





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