A new software company wants to start selling DVDs with their product. The manager notices that when the price for a DVD is 20 dollars, the company sells 139 units per week. When the price is 34 dollars, the number of DVDs sold decreases to 91 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:
A. To find the demand, q, as a function of price, p, we can use the given data points (20, 139) and (34, 91) to determine the slope of the demand curve.
First, calculate the slope using the formula:
slope = (change in q) / (change in p)
slope = (91 - 139) / (34 - 20)
slope = -48 / 14
slope = -3.43 (rounded to two decimal places)
Now we can use the slope-intercept form of a linear equation to find the equation of the demand curve:
q = m * p + b
where m is the slope and b is the y-intercept.
Using the point (20, 139), we can substitute the values into the equation to solve for b:
139 = -3.43 * 20 + b
139 = -68.6 + b
b = 207.6
Therefore, the equation for the demand, q, as a function of price, p, is:
q = -3.43p + 207.6
B. To write the revenue function as a function of price, we'll multiply the demand (q) by the price (p):
R(p) = p * q
R(p) = p * (-3.43p + 207.6)
Simplifying this expression, we have:
R(p) = -3.43p^2 + 207.6p
C. To find the price that maximizes revenue, we need to maximize the revenue function. One way to do this is by finding the vertex of the parabolic revenue function.
The revenue function, R(p), is in the form of a quadratic equation, -3.43p^2 + 207.6p. The vertex of a quadratic function is located at -b/2a, where a is the coefficient of the quadratic term and b is the coefficient of the linear term.
In this case, a = -3.43 and b = 207.6. Plugging these values into the formula:
p = -207.6 / (2 * -3.43)
p ≈ 30.28
Therefore, the price that maximizes revenue is approximately $30.
D. To find the maximum revenue, we can substitute the value of p from the previous question into the revenue function R(p):
R(p) = -3.43p^2 + 207.6p
R(30.28) = -3.43(30.28)^2 + 207.6(30.28)
R(30.28) ≈ 3029.71
Therefore, the maximum revenue is approximately $3029.71.