Let p be the price per hamburger and q be the number of hamburgers sold
Revenue = (unit price) x (quantity sold)
(40,000, 1.00) (20,000, 2.00)
(q, p) y = mx+b
slope = 1.00-2.00 = -1 40,000-20,000 20,000
1.00 = (-1/20,000)(40,000) + b
1.00 = -2 + b
+2 +2
3 = b y = (-1/20,000)x + 3 p = (-1/20,000)q +3 This is the demand equation
Revenue = q((-1/20,000)q + 3)
Revenue(q) = (-1/20,000)q^2 + 3q This is the revenue function
What is the increase in revenue as sales change from 20,000 to 20,001 hamburgers?
Revenue (q) = (-1/20,000)q^2 + 3q
Revenue (20,000) = (-1/20,000)(20,000)^2 + 3(20,000)
Revenue = $40,000
Revenue (q) = (-1/20,000)q^2 + 3q
Revenue (20,001) = (-1/20,001)(20,001)^2 + 3(20,001)
Revenue = $40,002
40,000-40,002 = $2 Increase in revenue
Determine the quantity and price that maximize revenue.
Revenue (q) = (-1/20,000)q^2 + 3q
A B
X = -B/(2*A) Q = -B/(2*A)
Q = -3/(2*(-1/20,000))
Q = 30,000 units of hamburgers will maximize revenue
To find max revenue = R(30,000)
R = (-1/20,000)(30,000)^2 + 3(30,000)
R = $45,000 is the maximum revenue
Price = 45,000/30,000 = $1.50 is the price that will maximize revenue
The cost of producing q hamburgers is C(q) = 5,000 +0.56q
Find the total profit for 20,000 units:
Rev (20,000) = (-1/20,000)(20,000)^2 + 3(20,000)
Rev = $40,000
Cost (20,000) = 5000 + 0.56(20,000) = $16,200
Profit = Rev – Cost
Profit = 40,000 – 16,200
Profit = $23,000 is the total profit for 20,000 units
Price = 40,000/20,000 = $2.00 is the price for 20,00 units
Find the total profit for 24,400 units:
Rev (24,400) = (-1/20,000)(24,400)^2 + 3(24,400)
Rev = $43,432
Cost (24,400) = 5000+ 0.56(24,000) = $18,664
Profit = Rev – Cost
Profit = 43,432 – 18,664
Profit = $24,768 is the total profit for 24,400
Price = 43,432/24,400 = $1.78 is the price for 24,400 units
Find