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I am stuck on the following question:

Consider a profit maximizing monopoly. The demand for the monopoly's product is given by $Q=\ln(a-bP)$ and its cost function is $C(Q)=ce^{Q}$, where the parameters $a$,$b$, and $c$ are all positive. Let $Q^{*}$ be the monopoly's optimal (profit-maximizing) output.

Derive an expression for $\frac{\partial Q^{*}}{\partial a} $ and determine its sign.

I know that profit ($\pi$) is $\pi=R(Q)-C(Q)$ (revenue less costs) where $R(Q)=PQ$. I need to find the max of $\pi(Q)$. First I need to write the demand function in terms of $Q$:

$$e^{Q}=a-bP$$ $$P=\frac{a-e^{Q}}{b} $$

Now I can derive the revenue function,

$$R=PQ=Q\left(\frac{a-e^{Q}}{b}\right) $$

and the profit function:


I now need to find where


$$\frac{a-e^{Q}(bc+Q+1)}{b} =0$$

I am unable to find the zeros. Perhaps I have not taken the right steps. Once I have found $Q^{*}$ (which is just the maximum point of $\pi (Q)$) I understand that I just take the partial derivative. Thank you for any help.

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up vote 1 down vote accepted

You don't need to find the zero to solve the problem. From your calculations we know that the optimum $Q^*(a)$ satisfies the equation

$$a- e^{Q^*} (bc + Q^{*} +1) = 0,$$

so differentiating with respect to $a$ gives

$$1 = \left(e^{Q^*} (bc+Q^{*} + 1) + e^{Q^*} \right) \frac{\partial Q^{*}}{\partial a}.$$

Thus $$\frac{\partial Q^{*}}{\partial a} = \frac{1}{a+e^{Q^*}}.$$

Since $a$ is positive, this expression must be positive.

EDIT: The punchline here, of course, is that you can derive useful information about the optimum -- increasing $a$ increases it -- even when actually calculating its value is unfeasibly complicated.

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