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, given that the demand curve is linear, we can use the two data points provided: (20, 139) and (34, 91).
We need to calculate the slope of the demand curve. The slope of a linear equation can be found using the formula: slope (m) = (change in y) / (change in x).
Using the first data point (20, 139), the change in x is (34 - 20) = 14 and the change in y is (91 - 139) = -48.
Therefore, the slope (m) = -48 / 14 = -3.4286 (approx).
Now we can find the equation of the line using the point-slope form: y - y1 = m(x - x1), where (x1, y1) is one of the given points.
Taking the point (20, 139) and substituting the values into the equation, we get:
q - 139 = -3.4286(p - 20)
Simplifying the equation, we get:
q = -3.4286p + 148.5714
So, the demand as a function of price is: q = -3.4286p + 148.5714.
B. To write the revenue function, we multiply the demand (q) by the price (p). Therefore, the revenue function is:
R(p) = p * q
Substituting the demand function we found earlier, we get:
R(p) = p * (-3.4286p + 148.5714)
Simplifying the equation, we get:
R(p) = -3.4286p^2 + 148.5714p
C. To find the price that maximizes revenue, we need to find the vertex of the revenue function graph. The vertex can be found using the equation:
p = -b / (2a), where a and b are the coefficients of the quadratic equation.
In our revenue function, a = -3.4286 and b = 148.5714. Plugging these values into the equation, we get:
p = -148.5714 / (2 * -3.4286) ≈ 21.76 (rounded to the nearest dollar)
Therefore, the price that maximizes revenue is approximately $22.
D. To find the maximum revenue, we substitute the price we obtained in the previous question into the revenue function:
R(p) = -3.4286(22)^2 + 148.5714(22)
Calculating the value, we get:
R(p) ≈ $1123.43
Therefore, the maximum revenue is approximately $1123.43.