wamap.org orums Calendar essment eek 10 Due in 1 hours, 32 minutes. Due Mon 06/1

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wamap.org orums Calendar essment eek 10 Due in 1 hours, 32 minutes. Due Mon 06/11/2018 Show intro/Instructions Score on last attempt:0 out of 1 score in gradebook: 0 out of 1 Re attempt last question below, or select another question hat holds 68000 spectators. With A baseball team plays in a stadium t been 29000. When the pric related to ticket price the ticket price at $9 the average e dropped to $7, the average attendence rose to 3 4000. Assume that attendence is linearly What ticket price would maximize revenue? s7.4 Points possible: 1 Unlimited attempts Score on last attempt: 0. Score in gradebook: 0 Submit ere to search

Explanation / Answer

Solution

Let p and n respectively represent the ticket price and the corresponding attendance.

Given, p and n are linearly related, let n = a + bp ………………………………….(1)

Further given, n = 29000 at p = 9 and n = 34000 at p = 7, we have from (1),

a + 9b = 29000 ………………………………………………………………………(2)

a + 7b = 34000 ………………………………………………………………………(3)

(2) - (3): 2b = - 5000 or b = - 2500…………………………………………………(4)

(4) in (3): a = 51500 …………………………………………………………………(5)

(4) and (5) in (1): n = 51500 – 2500p…………………………………………………..(6)

Now, revenue, R = np = 51500p – 2500p2 [from (6)]…………………………………(7)

To find p that maximizes R, equate dR/dp to zero and solve for p.

So, dR/dp = 51500 – 5000p = 0 or

p = 10.3   

Thus, revenue maximizing price is $10.3 ANSWER

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