Banana Computer Company sells Banana Computers both in the domestic and foreign markets. Because of the differences in the power supplies, a Banana computer purchased in one market cannot be used in the other market. This means that the company can use third degree price discrimination in order to maximize profits. Let’s suppose that it costs $1,000 to produce each computer (this is marginal and average cost). Let’s suppose further that the domestic and foreign demand curves are given as follows (the subscript "F" denotes "foreign" while the subscript "D" is used to denote "domestic"): PD=13,000 -20QD
PF= 17,000-40QF

Required:
a. What prices maximize profits for this firm? How many computers do they sell in each market? How much profit does the company earn?
b. Now, suppose that somebody figured out a wiring trick that allows a Banana computer built for either market to be costlessly converted so that it works in the other market. This destroys the company's ability to practice third degree price discrimination and forces them to charge the same price in both markets. What price maximizes the company's profits now? How many computers will they sell in each location? How much profit does the company earn?

with price discrimination

Domestic Price 7,000 Quantity 300

Profit (7,000 - 1,000) * 300 = 1,800,000

Foreing Price 9,000 Quantity 200

Profit (9,000 - 1,000) * 200 = 1,600,000

Total 1,600,000 + 1,800,000 = 3,400,000

no price discrimination:

Price 7,667 Quantity 500

Profit (7,667 - 1,000) x 500 = 3,333,500

Explanation:

Sales Revenue (Domestic)

We now equalice against Marginal Cost:

13,000 - 40Qd = 1,000

Qd = 12,000/40 = 300

Price: 13,000 - 20(300) = 7,000

We do the same process with Foreing demand:

(17,000 - 40Qf) x Qf = -40Qf^2 + 17,000Qf

R' = -80Qf + 17,000

-80Qf + 17,000 = 1,000

Qf = 16,000/80 = 200

Pf = 17,000 - 40(200) = 9,000

If the company cannot do price discrimination then:

We solve for the inverse of both market:

PD=13,000 -20QD

QD = 650 - PD/20

we take the price restrictions:

PD < 13,000

PF= 17,000-40QF

QF = (17,000 - PF)/40 = 425

QF = 425 - PF/40

PF < 17,000

Now, we aggregate the demands:

(650 -P/20 ) + (425 -P/40) =

Q= 1,075 - 0.075P

Make the inverse

P = (1,075 - Q ) / 0.075 = 14.333,33 -13.33Q

And solve for the Quantiy and Price that maximize profit

R = (14.333,33 -13.33Q) x Q = -13.33Q^2 + 14,333.33Q

R' = R(q)/dq = -26.66Q + 14,333.33

-26.66Q + 14,333.33 = 1,000

Q = 500

P = 14,333.33 - 13.33(500) = 7,667

i would say iii (3) because in order to compare a price you have to use as a reference other near stores because it's its

hope i explained myself