A retired auto mechanic hopes to open a rustproofing shop. Customers would be lo

Question

A retired auto mechanic hopes to open a rustproofing shop. Customers would be local new-car dealers. Two locations are being considered, one in the center of the city and one on the outskirts. The central city location would involve fixed monthly costs of $6, 880 and labor, materials, and transportation costs of $30 per car. The outside location would have fixed monthly costs of $4, 100 and labor, materials, and transportation costs of $40 per car. Dealer price at either location will be $90 per car. Which location will yield the greatest profit if monthly demand is 200 cars? 300 cars? At what volume of output will the two sites yield the same monthly profit?

Explanation / Answer

For the City Overhead cost(expenditure) = 6880$, Transportation cost/car(expenditure)= 30$ and Dealership cost(income) = 90$

For Outside the City Overhead cost(expenditure) =4100$, Transportation cost/car(expenditure)= 40$ and Dealership cost(income) = 90$

If we take the no of cars serviced/month to be = x, then we can form 2 profit equations, for cars serviced in the city and outside the city. Since Profit= Income - Expenditure

For the City, Profit = 90x-(6880+30x)----(1) and

For Outside, Profit = 90x-(4100+40x)----(2)

(a) (i) for 200 cars which will yield a greater profit

putting x=200 in equations (1)and(2)

Profit-city= 90*200-(6880+30*200)= 18000-(6880+6000)= 18000-12880= 5,120$

Profit-Outside city= 90*200-(4100+30*200)= 18000-(4100+6000)= 18000-10100= 7,900$

For 200 cars Outside the city is the better location

(ii) for 300 cars which will yield a greater profit

putting x=300 in equations (1)and(2)

Profit-city= 90*300-(6880+30*300)= 27000-(6880+9000)= 27000-15880= 11,120$

Profit-Outside city= 90*300-(4100+40*300)= 27000-(4100+12000)= 27000-16100= 10,900$

For 300 cars the city is the better location.

(b) At what volume of Output will the 2 sites yield the same monthly profit?

To get these we need to equate equations (1) & (2) and get the value of x

Profit-City= Profit-Outside city

i.e 90x-(6880+30x) = 90x-(4100+40x)-

therefore 4100 + 40x= 6880 + 30x

So 10x = 6880-4100 = 2780 or

x = 278 cars.

I.e at monthly volumes of 278 cars , both locations will yield the same amount of profit.

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