Check and correct my answers please. Thank you.

Trunks owns a junk yard. He can use one of two methods to destroy cars. The first method involves purchasing a hydraulic car smasher that costs $200/year to own and then 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 his brother, Scoop, to bury the cars at a cost of $5 each.

a) What are the cost functions for the two methods, where y is output per year?

b) What is the AC(y) and MC(y) of the first method and second method?

c) If Trunks wrecks 40 cars per year, which method should he use? If he wrecks 50 cars per year, which method should he use? What is the smallest number of cars per year for which it would be cost effective for him to buy hydraulic smasher (round to the nearest whole number)?

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my answers:

a)
c1(y) = y + 200
c2(y) = 5y + 10

b)
First Method:
AC(y)=1+200/y
MC(y)=$1 (I actually don’t know how to get this…just guessed the answer…)

Second method:
AC(y)=5+10/y
MC(y)=$5 (I actually don’t know how to get this…just guessed the answer…)

c)
Method 2
Method 1
25 cars per year (I actually don’t know how to get this…just guessed the answer…)

a) I agree

b1) I agree
b2) I agree
(To smash 1 additional car cost $1 in method1 and $5 in method2)
c1) I agree
c2) I agree
c3) I disagree. Set c1=c2 and solve for y. That is y+200=5y+10 (then round up)

a)

Your cost functions for the two methods are correct:
c1(y) = y + 200
c2(y) = 5y + 10

b)
For the first method:
AC(y) = (y + 200)/y
MC(y) = $1

For the second method:
AC(y) = (5y + 10)/y
MC(y) = $5

c)
To determine which method Trunks should use, we need to compare the costs for each method based on the number of cars wrecked per year.

If Trunks wrecks 40 cars per year:
Method 1: AC(40) = (40 + 200)/40 = $6.50 per car
Method 2: AC(40) = (5(40) + 10)/40 = $5.25 per car

In this case, Trunks should use Method 2 because it has a lower average cost per car.

If Trunks wrecks 50 cars per year:
Method 1: AC(50) = (50 + 200)/50 = $6 per car
Method 2: AC(50) = (5(50) + 10)/50 = $5.20 per car

Again, Method 2 has a lower average cost per car, so Trunks should use Method 2.

To find the smallest number of cars per year for which it would be cost-effective for Trunks to buy the hydraulic smasher, we can set the AC(y) for Method 2 equal to the AC(y) for Method 1 and solve for y:
(5y + 10)/y = (y + 200)/y
5y + 10 = y + 200
4y = 190
y ≈ 47.5

Rounding to the nearest whole number, the smallest number of cars per year for which it would be cost-effective for Trunks to buy the hydraulic smasher is 48 cars.

a) Your answers for the cost functions are correct.

The cost function for the first method is correct: c1(y) = y + 200. This includes the cost of owning the hydraulic car smasher ($200/year) and $1 for every car smashed into oblivion.

The cost function for the second method is also correct: c2(y) = 5y + 10. This includes the cost of purchasing a shovel ($10/year) and paying his brother, Scoop, to bury the cars at a cost of $5 each.

b) For the first method:
The average cost (AC) is calculated by dividing the total cost (c1(y)) by the output (y). So, AC(y) = (y + 200) / y.

The marginal cost (MC) is the additional cost incurred by producing one more unit. In this case, since each car smashed costs $1, the MC is constant and equal to $1.

For the second method:
The average cost (AC) is calculated by dividing the total cost (c2(y)) by the output (y). So, AC(y) = (5y + 10) / y.

Similarly, the marginal cost (MC) is constant and equal to $5, since each additional car being buried costs $5.

c) To determine which method Trunks should use for wrecking 40 cars per year and 50 cars per year, you should compare the total costs of each method for those outputs.

Method 1: c1(y) = y + 200
Method 2: c2(y) = 5y + 10

For 40 cars per year:
Method 1: c1(40) = 40 + 200 = $240
Method 2: c2(40) = (5 * 40) + 10 = $210

For 50 cars per year:
Method 1: c1(50) = 50 + 200 = $250
Method 2: c2(50) = (5 * 50) + 10 = $260

Therefore, for 40 cars per year, Trunks should use Method 2 (burying cars). However, for 50 cars per year, Method 1 (hydraulic car smasher) would be cheaper.

To find the smallest number of cars per year for which it would be cost-effective for Trunks to buy a hydraulic smasher, you need to compare the costs of the two methods. Set c1(y) = c2(y) and solve for y.

For Method 1: y + 200 = 5y + 10
Simplifying, 4y = 190
y ≈ 47.5

Rounded to the nearest whole number, the smallest number of cars per year for which it would be cost-effective to buy a hydraulic smasher is 48 cars per year.