(1 pt) A new software company wants to start selling DVDs with their
product. The manager notices that when the price for a DVD is 15
dollars, the company sells 132 units per week. When the price is
27 dollars, the number of DVDs sold decreases to 84 units per week.
Answer the following questions:
A. Assume that the demand curve is linear. Find the demand, q, as a function of price, p.
Answer: q=
B. Write the revenue function, as a function of price.
Answer: R(p)=
C. Find the price that maximizes revenue. Hint: you may sketch the
graph of the revenue function. Round your answer to the closest dollar.
Answer:
D. Find the maximum revenue, i.e., the revenue that corresponds to your answer for the preceding question.
Answer:
To find the demand, q, as a function of price, p, we can use the two points given: (15, 132) and (27, 84).
First, let's find the slope of the demand curve, which represents the rate at which the quantity demanded changes with respect to the price.
slope = (change in quantity) / (change in price)
= (84 - 132) / (27 - 15)
= -48 / 12
= -4
Next, we can use the point-slope form of a linear equation to find the equation of the demand function:
q - q1 = m(p - p1)
Using the first point (15, 132):
q - 132 = -4(p - 15)
q - 132 = -4p + 60
Rearranging the equation to solve for q:
q = -4p + 192
So, the demand, q, as a function of price, p, is:
q = -4p + 192
Moving on to part B, the revenue function, R(p), is the product of price and quantity demanded:
R(p) = p * q
Substituting the expression for q from part A:
R(p) = p * (-4p + 192)
= -4p^2 + 192p
For part C, to find the price that maximizes revenue, we need to find the vertex of the parabolic revenue function.
The revenue function is in the form of a quadratic equation, which has a vertex at p = -b / (2a), where a and b are the coefficients of the quadratic equation.
In this case, a = -4 and b = 192.
p = -192 / (2 * -4)
p = 192 / 8
p = 24
So, the price that maximizes revenue is $24 (rounded to the closest dollar).
For part D, to find the maximum revenue, we substitute the price into the revenue function:
R(p) = -4p^2 + 192p
R(24) = -4(24)^2 + 192(24)
= -2304 + 4608
= 2304
The maximum revenue is $2,304.