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Please explain or show how you got your answer, I need to learn this, not copy i

ID: 1187307 • Letter: P

Question

Please explain or show how you got your answer, I need to learn this, not copy it :)

A third brother, Rex Carr, owns a junk yard. Rex can use one of two methods to destroy cars. The first involves purchasing a hydraulic car smasher that costs $200 a year to own and the spending $1 for every car smashed into oblivion; the second method involves purchasing a shovel that will last one year and costs $10 and paying the last Carr brother, Scoop, to bury the cars at a cost of $5 each.

a) write down the total cost functions for the two methods, where y is output per year: TC1 (y)=_____, TC2 (y)=________

b) the first method has an average cost function_______ and a marginal cost function______. For the second method these costs are ______ and _______.

c) If Rex wrecks 40 cars per year, which method should he use?  If Rex wrecks 50 cars per year, which method should he use? What is the smallest number of cars oper year for which it would pay him to buy the hydraulic smasher?

Explanation / Answer

a) think about the regular algebric problem, initial cost is 200, and then it cost $1 every smash. For second one initial cost is $10, and costs $5 to bury


TC1 = 200 + y TC2 = 10 + 5y


b) for average cost you just divide by total output (y)


(200 + y) / y = 200/y + 1..................marginal cost = 1

(10 + 5y)/y = 10/y + 5......................marginal cost = 5



c) plug in 40, and 50 in answer (a) and see which one costs less

method 1

40 cars: 200 + 40 = 240

50 cars: 10 + 5*40 = 210


method 1

40 cars: 200 + 50 = 250

50 cars: 10 + 5*50 = 260


for 40 cars method 2 is better, and for 50 cars method 1 is better.


for secon part of (c), overall hydraulic cost starts high becasue initial cost is 200 much higher than method 2 burying. So its asking you to find number of cars when it will both method will be equal.

so


200 + y = 10 + 5y

200 - 10 = 5y - y

190 = 4y

y = 47.5


so it will need 48 cars to make method 1 better than method 2.