The organizer of a sports event estimates that if the event is announced ๐ฅ days in advance, the revenueobtained will be ๐ (๐ฅ) thousand dollars, where๐ (๐ฅ) = 400 + 120๐ฅ โ ๐ฅ2The cost of advertising the event for ๐ฅ days is ๐ถ(๐ฅ) = 2๐ฅ2 + 300a. Find the profit function ๐(๐ฅ) = ๐ (๐ฅ) โ ๐ถ(๐ฅ) and sketch its graph.b. How many days in advance should the event be announced in order to maximize profit? What is themaximum profit?c. What is the ratio of revenue to cost ๐(๐ฅ) = ๐ (๐ฅ)๐ถ(๐ฅ) at the optimal announcement time found in part (b)?What happens to this ratio as ๐ฅ โ 0? Interpret these results
Question
The organizer of a sports event estimates that if the event is announced ๐ฅ days in advance, the revenueobtained will be ๐ (๐ฅ) thousand dollars, where๐ (๐ฅ) = 400 + 120๐ฅ โ ๐ฅ2The cost of advertising the event for ๐ฅ days is ๐ถ(๐ฅ) = 2๐ฅ2 + 300a. Find the profit function ๐(๐ฅ) = ๐ (๐ฅ) โ ๐ถ(๐ฅ) and sketch its graph.b. How many days in advance should the event be announced in order to maximize profit? What is themaximum profit?c. What is the ratio of revenue to cost ๐(๐ฅ) = ๐ (๐ฅ)๐ถ(๐ฅ) at the optimal announcement time found in part (b)?What happens to this ratio as ๐ฅ โ 0? Interpret these results
Solution
a. The profit function P(x) is obtained by subtracting the cost function C(x) from the revenue function R(x). So, P(x) = R(x) - C(x) = (400 + 120x - x^2) - (2x^2 + 300). Simplifying this gives P(x) = 100 + 120x - 3x^2.
b. To find the number of days in advance the event should be announced to maximize profit, we need to find the maximum of the profit function. This occurs at the vertex of the parabola represented by the profit function. The x-coordinate of the vertex of a parabola given by y = ax^2 + bx + c is -b/2a. So, the number of days in advance to announce the event is -120/(2*-3) = 20 days. Substituting x = 20 into the profit function gives the maximum profit: P(20) = 100 + 12020 - 320^2 = $2200.
c. The ratio of revenue to cost Q(x) = R(x)/C(x) at the optimal announcement time is Q(20) = R(20)/C(20) = (400 + 12020 - 20^2) / (220^2 + 300) = 1.22. As x โ 0, Q(x) โ R(0)/C(0) = 400/300 = 4/3. This means that as the announcement time approaches 0, the revenue is 4/3 times the cost.
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