# Maximising profit

• September 13th 2008, 10:00 PM
brumby_3
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?
• September 14th 2008, 06:50 AM
CaptainBlack
Quote:

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:

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

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
• September 14th 2008, 01:18 PM
brumby_3
Can someone further step-by-step show me how to solve this problem so I can try for some other questions I have please? (Cool)
• September 14th 2008, 01:58 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Profit:

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

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