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I'm having trouble on my homework and I need some help. A company sells products in a perfectly competitive market, where the price is $p = 24.$ For each of the following cost functions, write down the company’s profit-maximization problem, and then find $q^∗$.

i. $C(q) = 2q^3$

ii. $C(q)=30q+q^2$

iii. $C(q) = 10q$

iv. $C(q) = 24q$

I finished i and ii on my own, for i, $$\max_q \text{Total Revenue}-\text{Total Cost} = \max_q 24q-2q^3,$$ then I take the derivative of max, I get $24-6(q^*)^2$ and set it equal to 0, I get $q^*=2$ and $q^*=-2$, then use the second-order condition find out $q^*=2$. For ii, I get $q^*=-3$. But for iii, I get $$24q-10q=14q,$$ when I take derivative I get $14$, how can I set $14=0$? What is $q^*4$ in this situation? Or does $q^*$ even exist? Also for iv, $max'$ is $0$, what is $q^*$ in this situation? Any help would be appreciated.

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My professor once said, when doing economics, don't get stuck in mathematics. Math is just a tool.

You know that the price will always be 24 per piece. For (iii), your cost is $C(q) = 10q$. What's the cost per piece to produce it? Is it more or less than what you could sell? If it's the former, you're guaranteed to make profit for each piece you make. If it's the latter, you'll make a loss for each piece you make. Given this, what does that say about the quantity you should produce?

Edit: well you could get stuck in math... but need to realize that for a function $f(x)$ to be maximized at $x^*$ where $f'(x^*) = 0$ you need the second order condition, namely $f''(x^*) < 0$, to hold as well.

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You can get some insight by looking at iv). It's a similar problem, but has a very clear and easy to see result. These questions are helping you remember that "corner solutions" exist, and you should always be on the lookout for them.

Remember that you can't find a unique solution to the optimization problem if your objective function (profits = revenues - costs) has no critical points. This is OK - in the real world, countless frictions exist that are not expressed in your model, but would probably take care of any unintuitive results.

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  • $\begingroup$ Strictly speaking, corner solutions are often unique, but I didn't want to get too nuanced here because there uniqueness usually comes from additional constraints to the problem (e.g., x >= 0). $\endgroup$ – heh Oct 3 at 16:58

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