Thread: Urgent. Cost demand functions

The latest demand equation for your Yoda vs. Alien T-shirts is given by the formula below, where q is the number of shirts you can sell in one week if you charge $x per shirt. q = -50x + 700
The Student Council charges you $100 per week for use of their facilities, and the T-shirts cost you $6 each. Find the weekly cost as a function of the unit price x.
C(x) =?
Hence, find the weekly profit as a function of x.
P(x) =

per week:
demand, q = -50x +700-----if selling price is x per shirt.

Total Cost, C = 100 +6*q
C(x) = 100 +6[-50x +700]
C(x) = 100 -300x +4200
C(x) = -300x +4300 ----------------answer.

Revenue = x*q
R(x) = x[-50x +700] = -50x^2 +700x

Profit = Revenue minus Cost
P(x) = R(x) -C(x)
P(x) = [-50x^2 +700x] -[-300x +4300]
P(x) = -50x^2 +700x +300x -4300
P(x) = -50x^2 +1000x -4300 ---------------answer.

---------------
q = -50x +700
If no demand, q=0,
50x = 700
x = 14
That means x < 14, to have some demand.

Say x = 13
q = -50(13) +700 = 50 shirts
P(13) = -50(13^2) +1000(13) -4300
P(13) = 250

Say x = 2(6) = 12
q = -50(12) +700 = 100 shirts
P(12) = -50(12^2) +1000(12) -4300
P(12) = 500

Say x = 11
q = -50(11) +700 = 150 shirts
P(11) = -50(11^2) +1000(11) -4300
P(11) = 650

Say x = 10
q = -50(10) +700 = 200 shirts
P(10) = -50(10^2) +1000(10) -4300
P(10) = 700

Say x = 9
q = -50(9) +700 = 250 shirts
P(9) = -50(9^2) +1000(9) -4300
P(9) = 650

Going down. So max profit is when price is 10 per shirt.
This is not asked. I'm only playing.