Please answer the question 4-5. Suppose that there is only one newspaper publish
ID: 1201439 • Letter: P
Question
Please answer the question 4-5.
Suppose that there is only one newspaper published in a small town. The demand for the paper depends on the price and the amount of news reported. The demand function is D(s,p) = 15 squareroot s/p^3 where s is the amount of news and p is the price. The cost of collecting news incurs cost: the cost to write s amount of news is 10s and this cost is independent the number of papers sold. There is additional cost to print and deliver the paper, which is given by $0.10 per copy and this is again independent of the amount of news. Therefore, the total cost of printing y copies of the paper with s amount of news is 10s + 0.10y. Calculate the price elasticity of demand for this newspaper. Does this price elasticity depend on the amount of news? Is it constant over all prices? We, in class, derive that MR(p) = p(1 + 1/epsilon). To maximize profits, the newspaper will set MR equal to MC. Solve for the profit maximizing price for the newspaper to charge per newspaper. If the newspaper charges the profit-maximizing price and writes 100 news, how many copies would it sell? Derive the number of copies sold as a function of s. Assuming that the paper charges the profit maximizing price, derive the profit as a function of y (without using the demand curve yet) and s. Using solution for y(s) that you found in (3), substitute y(s) for y and derive the profit a s a function of s alone. If the newspaper charges its profit-maximizing price and writes the profit-maximizing amount of news, how many news does it write? Given the profit-maximizing amount of news you calculated, how many copies are sold? What is the amount of profit for the newspaper?Explanation / Answer
4)
Given the elasticity of -3, the MR=MC rule suggests a price of $0.15 per copy. At this price, the quantity of newspaper sold is given by:
y = 15(s)0.5/(0.15)3
= 40000(s)0.5/9
Profit is the difference of total revenue and total cost
Pr = Py – 10s – 0.10y
= 0.15*40000(s)0.5/9 – 10s – 0.10y
= 6000(s)0.5/9 – 10s – 0.10*40000(s)0.5/9……. Substitute y = 40000(s)0.5/9
= 2000(s)0.5/9 – 10s
Profit is now a function of amount of news alone.
5) Profit is maximum at the level of s* where its derivative is zero
(0.5)*2000(s)-0.5/9 – 10 = 0
1000/9(s)-0.5 = 10
(s)0.5 = 100/9
s = 10000/81
The optimum amount of news is thus 10000/81.
At this amount, the number of copies sold is:
y = 15*(100/9)/(0.15)3
= 4,000,000/81
Profit = 2000/3*100/9 - 10*10000/81
= 500000/81