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monopolistic firm involving income effect and substitution effect

Given the demand function of a monopolist operating two plants is equal to Q=400-4p and cost functions for plant A is equal to C=20Q and plant B is equal to C=0.5Q²
Assuming the monopolist can price discriminate determine the profit maximizing output from plant A and from plant B and the price charged by the monopolist.
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2 Answers

Let Q1 and Q2 be the outputs of plants A and B. Solve the demand function for price and find total revenue:
 
P=100 - Q/4
TR = PQ = 100Q - Q²/4
TR = 100(Q1+Q2) - (Q1+Q2)²/4
 
Find total cost:
 
TC = 20Q1+0.5Q22
 
Profit = total revenue minus total cost:
 
Π = TR - TC = 100(Q1+Q2) - (Q1+Q2)²/4 - 20Q1 - 0.5Q22
 
Now take the partial derivatives and set them to zero to find the critical point(s):
 
Π1 = 100 - (Q1+Q2)/2 - 20 = 0
Π2 = 100 - (Q1+Q2)/2 - Q2 = 0
 
Solving these two equations gives only one critical point:
 
Q1=140,  Q2=20.
 
You can check that this critical point is indeed a maximum of the profit function by checking the 2nd partials:
 
Π11 = -1/2 < 0, Π22 = -3/2 < 0, Π11Π22 - Π122 = 1/2 > 0.
 
These are the conditions for a maximum.
 
Therefore, the total output of the two plants is Q=140+20=160, at a price of P=100-160/4=60.
 
 
econ answers

The problem can be solved in two steps.

I'm assuming you're class is calculus-based; if not, you can figure out the
maximum profit by plotting it, by finding the maximum of the parabola (you
might need to remember how to complete the square from Algebra 2 or
precalculus), or by calculating a lot of production values and hoping to
stumble across the right one (not recommended).

Whether it's calculus-based or not, and whether it's a monopoly or not, the
firm will produce until marginal revenue equals marginal cost.

The marginal revenue can be calculated by calculating the derivative of
total revenue. Total revenue (TR) always equals price times quantity.
Quantity is your independent variable, so it's handy to invert the given
equation to get price as a function of quantity.

Q = 400-4P

Q-400 = -4P

P = 100-0.25Q.

Total Revenue = PQ

= (100-0.25Q)*Q

= 100Q-0.25Q^2

Marginal revenue = d(TR)/dQ

= 100-0.5Q.

Calculating the cost function is a bit trickier, given that you have two
different factories. If you graph it, or plug in values, you will find that
the costs are initially lower at plant B. But at a certain point, plant B
becomes too expensive and production should shift to plant A.

You can find that critical point by setting the marginal costs equal to each
other, and solving for Q. You can find the marginal cost by taking the
derivative with respect to Q of both cost functions.

Marginal cost for factory A = 20
Marginal cost for factory B = Q

Obviously, when the marginal costs are equal, Q = 20. This means that for
Q<20, it's cheaper to produce all units at factory B. For Q>20, you should
produce 20 units at factory B, then produce the remaining units at factory
A.

So if the Q that maximizes profit is less than or equal to 20, then our cost
function is C = 0.5Q^2.

Then the marginal cost function would be MC = Q.

If the Q that maximizes profit is greater than 40, then the cost function
will be

C = 0.5(20)^2 +20(Q)

C = 200 + 20 Q

and the marginal cost function would be MC = 20.

Now what? Well, remember the condition for maximum profit is to set marginal revenue equal to marginal cost. But which marginal cost function? Well, the best way to do this is to actually set it equal to both, then check to see if the answer achieved is self-consistent. Then, if there are still multiple possible answers, we would need to calculate the profit function and substitute the possible answers.

This is a bit more complicated than if you have a single profit function. But that's probably why your teacher assigned it.

So, if Q<=20, then

MR = MC

100-0.5Q = Q

100 = 1.5Q

Q = 66.7.

This is inconsistent with the assumption that Q<=20. This means we should NOT use this marginal cost function, and instead use the one for Q>20.

If Q > 20, then

MR = MC

100-0.5Q = 20

-0.5Q = -80

Q = 160.

Therefore, the profit-maximizing output is Q = 160 units.