Please explain or show how you got your answer, I need to learn this, not copy i
ID: 1187306 • 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? Â 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)Here, in the first case, we have a fixed cost of $200 i.e. we have to pay the $200 irrespective of how many cars we break and then we have a variable cost that will depend on the number of cars that is smashed each year.
So, TC1(y)= 200+y
Similarly, in this case, we have a fixed cost of $10 i.e. we have to pay the $10 irrespective of how many cars we bury and then we have a variable cost that will be five times the number of cars that is buried each year.
So, TC2(y)=10+5y
b)
Average cost is equal to total cost divided by the number of goods produced (the output quantity, Q). And marginal cost is the change in the total cost that arises when the quantity produced changes by one unit. That is, it is the cost of producing one more unit of a good.
So, for average cost for TC1(y)=(200/y)+1 and the marginal cost for TC1(y)=dTC2/dy =d(200+y)/dy =1
So, for average cost for TC1(y)=(10/y)+5 and the marginal cost for TC2(y)=dTC2/dy =d(10+5y)/dy =5
For 40 cars TC1(y)= 200+40x1 =240 and TC2(y)=10+5x40=210. So, he should go for TC2.
For 50 cars TC1(y)= 200+50x1 =250 and TC2(y)=10+5x50=260. So, he should go for TC1.
the number of cars for which both the methods woud have equal cost=>
=>200+y=10+5y
=>y=190/4=47.5
As, y has to be whole number, we take the just bigger whole number i.e. 48. So, y=48