well, recall what you were told:
The price(in dollars) p,and the quantity sold x obey the demand equation
p=-1/10x+150
You found that maximum revenue R occurs at x=750, so
p = -75+150 = 75
Demand Equation. The price(in dollars),and the quantity x sold of a certain product, obey the demand equation. I can't figure out what price the company should charge to maximize revenue
p=-1/10x+150
R=p*x
R=((-1/10)*x+150)*x
R=(-1/10)*x^2+150x
y=ax^2+b*x+c
a=-1/10
b=150
c=0
-2 b/2a=150/2*(-1/10)
-b/2a=150/1/5
-b/2a=750
R max=(-1/10)*x^2+150x
R max=(-1/10)*750^2+150*750
R max=-562500/10+112500
R max=-56250+112500
R max=56250
1 answer
1 answer