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I'm reviewing the following question from my microeconomics course notes. I don't understand how to do part (b).

A firm faces a perfectly elastic demand curve because the product it sells has a perfect substitute. The (inverse) demand function is $p=2$. Its production function is $f(K,L)=K^{1/4}L^{1/4}$.

(a) Assuming input prices of $w=r=1$, show that the firm's cost function is $C(q)=2q^2$.

(b) Assuming the firm operates as a monopolist, what price will it charge, what quantity will it produce, and what are its profits?

(c) What is the deadweight loss from monopoly?

(d) Would the answers to parts (b) and (c) change if the firm could perfectly price discriminate?

Part (a) is easy; I'm stuck on part (b).

If the firm faces a "perfectly elastic demand curve" $p=2$, then the firm isn't in a position to operate as a monopolist, right? A monopolist ought to be able to set a price, not have to accept the market price $p=2$.

If I assume the monopolist sets a price $p$ for quantity supplied $q$, then I can define the profit function

$$\pi(q)=p(q)\cdot q-c(q)=p(q)\cdot q-2q^2$$

Differentiating gives the first-order condition

$$q\cdot p'(q)+p(q)=4q$$

But I can't proceed without knowing the monopolist's demand curve (which, again, I thought couldn't be perfectly inelastic).

If I assume $p(q)=2$, then I can deduce that $q=1/2$. But this can't be right, because then I can't calculate a deadweight loss.

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  • $\begingroup$ If the price is effectively fixed, then I would agree that (c) and (d) are rather empty questions: price is equal to marginal revenue as no customer is willing to pay more than $2$, but enough customers are willing to pay any amount less than or equal to $2$ for more than the firm can produce $\endgroup$ – Henry Aug 20 '17 at 13:29
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The monopolist is allowed to set the price. However, because his product has a perfect substitute, he can sell 0 products at prices $p>2$. He also has no incentive seting the price $p<2$, because the consumers will buy any number of products he wishes to sell at $p = 2$. This means that $p(q) = 2$. Its graph in the $(q,p)$ coordinate system would be a horizontal line at $p = 2$. Because of this, $p'(q) = 0$. Correcting for this you should be able to easily finish the exercise.

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