Open stevecondylios opened 4 years ago
Here's a simple example (in this example, we assume a constant cost function):
profit (π) qp - qc = q(p - c) ..........................................................................(1)
demand q = 200 - 20p ...................................................................................(2)
(2) into (1)
(200 - 20p)(p - c) = 200p - 200c - 20p^2 + 20pc ........................................................(3)
This gives profit for any given price
To find the price that maximises profit, take partial derivative w.r.t price:
∂π/∂p = 200 - 40p + 20c ...............................................................................(4)
We must provide the cost function, in this case we assume a constant cost function:
c = $4 .................................................................................................(5)
(5) into (4)
∂π/∂p = 200 - 40p + 80 = -40p + 280 ....................................................................(6)
Set to zero to optimise
-40p + 280 = 0
p = $7
Which is the profit maximising price for a given demand curve/equation and unit cost
Function should return pmax price, as well as π
A simple functional to accept a demand curve function, cost function, and return π max.
Experiment with including parameters for some 'degree' of π max (e.g. 30%, 60%, 90%) etc.