Answer to Question #164823 in Microeconomics for Leyton Juma Hosea

Question

Given this information for a monopoly company producing cars :

1. cost of production : 2 million ( 2000 in thousands) w/ nothing to maintain 
2. foreseeable demand : q= 900-150p ( q is the number of drivers (thousands) given the price per driver p)

What would be the profit-maximizing problem (optimization)

What would be the profit-maximizing quantity and price?

Answer 1

Solution:

Profit maximizing problem is given by:

P(q) = R(q) – C(q)

Where: P(q) = Profit

R(q) = Total Revenue (TR)

C(q) = Total Cost (TC)

Given the demand function: q = 900 – 150p, compute the inverse demand function by solving for p in the demand function.

q = 900 – 150p

p = $(\frac{q-900 }{150} )$ = 0.0067q – 6

Find TR = $P\times Q$

=$(0.0067q - 6) (Q)$

= $0.0067Q^{2} -6Q$ 0.0067Q² – 6Q

Derive the Marginal Revenue (MR) by getting the derivative of the TR function:

MR = 0.0134Q – 6

Find the Total Cost (TC) function = C x Q

= $(2,000,000) (Q)$ = 2,000,000Q

Derive the Marginal Cost (MC) by getting the derivative of the TC function:

MC = 2,000,000

Profit Maximization is where: MR = MC

= 0.0134Q – 6 = 2,000,000

= 0.0134Q = 2,000,000 + 6

= 0.0134Q = 2,000,006

Q =$\frac{2,000,006}{0.0134}$

Q = 149,254,179

The profit maximizing quantity is = 149,254,179 units

Find P = 0.0067Q – 6

Substitute for Q:

= $(0.0067\times 149,254,178) - 6$

= 1000,003 – 6

P = 999,997

The profit-maximizing price = 999,997