Question

Question 1

A local government board hired you to recommend which of the two possible sites should be chosen for a new recycling facility. Location A is much closer to the city center, and thus the average hauling distance would only be 2 miles, while Location B being outside of the city would result in an average hauling distance of 5 miles. Public funds will have to be used to pay for the leasing cost of the site, which would be $50,000 a year for Location A and $10,000 a year for Location B, as well as for the hauling cost which is $200 per mile for each trip. Assuming 3,000 trips will be made per year, which location should be chosen based on the overall cost?

Question 2 :

Metal ABC produces sheets of metal. Its fixed cost in year 2019 is $1,000,000. The variable cost is $400 per sheet. a. Find the break-even quantity assuming the selling price is $500 per sheet. b. Find the price and quantity that will result in maximum profit, assuming the demand function is ? = $10,000 − 6?. How much profit will be made using the profit-maximizing price and quantity values?

Answer 1

Answer 1

Location A

Leasing cost =$50,000 per year

Hauling cost = No of trips * Distance* cost per trip per mile = 3000*2*200=1,200,000

Total Cost = 50000+1200000=$1,250,000

Location B

Leasing cost =$10,000 per year

Hauling cost = No of trips * Distance* cost per trip per mile = 3000*2*500=3,000,000

Total Cost = 10000+3000000=$3,010,000

Hence Location A is Chosen

Answer 2

Let the break even quantity be x

Then Variable cost = 400*x =400x

Revenue = 500*x=500x

At break even, Revenue = Fixed cost+ variable cost

Hence, 500x=400x+1000000

or, 500x-400x=1000000

or, 100x =1000000

or, x = 1000000/100=10000 units

Hence break even units is 10000

For Q units, P=10000-6Q

Revenue = (10000-6Q)*Q=10000Q-6Q²

Cost = 1000000+400Q

Hence Profit = Revenue-Cost = 10000Q-6Q² -1000000-400Q = -1000000+9600Q-6Q²

For maximum Profit dP/dQ=0

Hence  d/dQ(-1000000+9600Q-6Q²)=0

or, 9600-6*2Q=0

or, 12Q=9600

or, Q=9600/12=800 units

P=10000-6Q = 10000-6*800 =10000-4800=$5200

Hence Profit maximization Q=800 units and P=$5200

Profit = -1000000+9600*800-6*800² = -1000000+7680000-6*640000=6680000-3840000=$2840000