A mining company extracts metal ore worth $A of revenue in year 0. Its costs per

Question

A mining company extracts metal ore worth $A of revenue in year 0. Its costs per year are $K, a constant. But the company expects its revenue to decline by p% each year. Its mining activity will continue as long as the yearly revenue exceeds the yearly costs K.
(a) What is the revenue in year n? For how many years n* does the company operate?

(b) Find the total profit in the period from year 0 to n*.

(c)Put A=7000000, K=5000000, and p=2%. (i)What is n* in this case?
(ii) What is the associated total profit? (if you cannot solve (c)(i), pick arbitrarily n* = 14.)

Hint: For any real number x is not equals to 1 and any ( finite) integer m, we have: m to the i=0 xi = (1- xm+1) / (1-x) .

Explanation / Answer

(a) Cost in all the years is K

Revenue is Year 0 = A ; Revenue in year 1 = A - (p/100) {as it decreases by p% every year);

Revenue in year 2 = Revenue in year 1 - p% = A - (p/100) - (p/100) = A - (2p/100)

Similarly revenue in year 3 = A - (3p/100)

Therefore Revenue in year n would be = A - (np/100)

The company will operate till yearly revenue exceeds cost so

A - (np/100) > =K

= A - K > = np/100

= np < =100 (A-K)

= n < = 100(A-K)/p

Therefore n* = 100(A-K)/p

(b) Total profit from year 0 to year n*

Profit in year 0 = A-K

Profit in year 1 = A - (p/100) - K

Profit in year 2 = A - (2p/100) - K

Profit in year n* = A - 100(A-K)/p*(p/100) - K = A - A + K - K = 0 (which ideally should be)

So total profit would be n*A - (p/100 + 2p/100 + 3p/100.........) - n*K

(c)  A=7000000, K=5000000, and p=2%

c(i) n * = 100(A-K)/p = 100 * (7000000 - 5000000)/2 = 100000000 = 100 million years

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