Pages: 4 (628 words) Published: March 1, 2013
Find the Revenue Function:

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 +3This is the demand equation

Revenue = q((-1/20,000)q + 3)
Revenue(q) = (-1/20,000)q^2 + 3qThis 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

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