Question

Profit-maximizing Q (quantity) and P (price)

will you get a different Q and P if you use equations 2 and 4 vs. equations 2, 3, and 5?

(1) Demand: Q = 230 – 2.5P + 4*Ps + .5*I, where Ps = 2.5, I = 20.

(2) Inverse demand function [P=f(Q)], holding other factors (Ps = 2.5 and I =20) constant, is, P=100-.4*Q.

(3) Production: Q = 1.2*L - .004L2 + 4*K - .002K2;

(4) Long Run Total Cost: LRTC = 2.46*Q + .00025*Q2 (Note: there are no Fixed Costs);

(5) Total Cost: TC = 1*L + 10*K.

Answer 1

Let us use equations 2 and 4:

Profit = PQ - LRTC = PQ - 2.46Q - 0.00025Q²

Profit = 100Q - 0.4Q² - 2.46Q - 0.00025Q²

Profit = 97.54Q - 0.40025Q²

Differentiating wrt Q and equating it to zero we get:

d(Profit)/dQ = 97.54 - 0.8005Q = 0

Q = 97.54/0.8005 = 121.84884

P = 100 - 0.4Q = 100 - 0.4(121.84884) = 51.26046

P* = 51.26046 Q* = 121.84884

Now let us solve using the equations 2,3 and 5:

Profit = PQ - TC

Profit = 100Q - 0.4Q² - L - 10K

Q here is a function of L and K. Q = 1.2L - 0.004L² + 4K - 0.002K²

Now for profit maximisation, we need to differentiate Profit wrt L and K and equate both to zero.

d(Profit)/dL = 100dQ/dL - 0.8QdQ/dL - 1 = 0

Now, dQ/dL = 1.2 - 0.008L

d(Profit)/dL = (100 - 0.8Q)(1.2 - 0.008L) - 1 = 0

Hence we get:

(100-0.8Q)(1.2 - 0.008L) = 1 Let this be equation 1

d(Profit)/dK = (100 - 0.8Q)(dQ/dK) - 10 = 0

dQ/dK = 4 - 0.004K

d(Profit)/dK = (100 - 0.8Q)(4-0.004K) - 10 = 0

Hence we get:

(100 - 0.8Q)(4-0.004K) = 10 Let this be equation 2

Dividng equation 2 by equation 1 we get:

(4-0.004K) = 10(1.2 - 0.008L)

4 - 0.004K = 12 - 0.08L

Hence we get;

K = ((0.08L - 8) / 0.004)

K = 20L - 2000 = 20(L - 100)

Hence we substitute this relation in the production function, we get:

Q = 1.2L - 0.004L² + 4(20(L - 100)) -0.002(20(L-100))²

Q = 1.2L - 0.004L² + 80L - 8000 - 0.8L² + 160L - 8000

Q = 241.2L - 0.804L² -16000

Using this relation in equation 1 we get:

(100-0.8Q)(1.2-0.008L) = 1

(100 - 192.96L + 0.6432L² +12800)(1.2 - 0.008L) = 1

Hence we get a cubic equation

Using a calculator we get:

L = 100.563, 150, 199.437

Let us find the Q for each case:

1) For L = 100.563:

Q = 241.2L - 0.804L² -16000 = 125.01036

2) For L = 150

Q = 241.2L - 0.804L² -16000 = 2090

3) For L = 199.437

Q = 241.2L - 0.804L² -16000 = 125.01036

Hence we can see that for the first and the last case, Q is very close to the one obtained if we were using equations 2 and 4. Hence we can say that we will get a different Q and P if we were using equations 2 and 4 vs. equations 2,3 and 5 but they will only be marginally different(i.e very less difference)