PLEASE HELP ME!! A boat leaves a dock at 1:00 PM and travels due south at a spee
ID: 3346242 • Letter: P
Question
PLEASE HELP ME!!
A boat leaves a dock at 1:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 2:00 PM. How many minutes after 1:00 PM were the two boats closest together? (Round your answer to the nearest minute.) A baseball team plays in a stadium that holds 59,000 spectators. With ticket prices at $10, the average attendance had been 5,000. When ticket prices were lowered to $8, the average attendance rose to 15,000. Find the demand function (price p as a function of attendance x), assuming it to be linear. P(x) = How should ticket prices be set to maximize revenue? (Round your answer to the nearest cent.)Explanation / Answer
1. Let (0, 0) denote the first boat at 1pm. At time t, it is located at (0, -20t). Since the second boat reaches the dock an hour later, it is located at (-15, 0) at 1pm. It heads east at 15km/h and so is located at (-15 + 15t, 0) at time t. The distance between these two are: D = ?[(0 + 20t)^2 + (-15 + 15t - 0)^2] = ?(625t^2 - 450t + 225) Then, by taking derivatives: D' = (1250t - 450)/[2?(625t^2 - 450t + 225)] Setting this equal to zero: (1250t - 450)/[2?(625t^2 - 450t + 225)] = 0 ==> 1250t - 450 = 0 ==> 1250t = 450 ==> t = 0.36 Therefore, the distance is minimized 0.36 hours * 60 minutes/1 hour = 22 minutes after 1pm. 2. Let p=price and y=attendance The two (x,y) coordinates are (10,5000) (8,15000) The linear equation using these two points is x1= 10 y1= 5000 x2= 8 y2= 51000 slope = (y2-y1)/(x2-x1) = (15000-(5000))/(8-(10)) = (15000-5000)/(8-10) slope= -5000 b = (y1-m*x1) b =5000-[(2000)/(-2)](10) = 5000 Equation: y= mx +b , m = slope Equation of the demand function is : p = -5000 x+ 5000 b) Revenue = price x tickets sold R=px R=(-5000x+5000)x R=-5000x^2+5000x dR/dx =-2000x+5000=0 solve for x x=5000/2000=2.5 Price $2.50 per ticket to maximize revenue not sure about second one but first one is correct
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