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$p= 10q^{-0,5}$
$C(q) = 5q$

In a monopoly the equation for maximum profits is: $p'(q)\cdot q + p(q) - C'(q) = 0$

First order condition:
$-0,5 \cdot 10q^{-1,5} \cdot q + 10q^{-0,5} - q = 0$

can't figure out onwards from here. How do I do with these negative exponents?

the answer is $q = 1$, and $p = 10$

Em.
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Hellin
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  • the answer is q = 1, and p = 10 btw – Hellin Aug 27 '16 at 18:48
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    Please review my edit. I suspect that is what you meant, but not everyone uses commas , to mark decimal places. Formatting tips here. – Em. Aug 27 '16 at 19:06
  • I think you may have differentiated $C(q)$ wrong. The equation should come out to be $(-.5)\cdot 10q^{-1.5}\cdot q + 10q^{-.5} - 5 = 0$ – Sean Haight Aug 27 '16 at 19:15

1 Answers1

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As stated above in a comment I believe the correct equation for monopoly profit optimization should be $(-.5)\cdot 10q^{-1.5}\cdot q + 10q^{-.5} - 5 = 0$. We will solve this equation for $q$ below:

$(-.5)\cdot 10q^{-1.5}\cdot q + 10q^{-.5} - 5 = 0 \Rightarrow \frac{-5}{\sqrt{q}} + \frac{10}{\sqrt{q}} - 5 = 0$

Here we have simply used algebra of exponents: $x^{-a} = \frac{1}{x^a}$ and $(x^a)(x^b) = x^{a+b}$.

$\frac{-5}{\sqrt{q}} + \frac{10}{\sqrt{q}} - 5 = 0 \Rightarrow \frac{5}{\sqrt{q}} = 5 \Rightarrow \sqrt{q} = 1 \Rightarrow q = 1$

Thus we arrive at $q = 1$. Plugging $q$ into our equation for $p$ we see $p = 10(q)^{-.5} = 10(1)^{-.5} = 10(1) = 10$. Thus $q = 1$ and $p = 10$.

Sean Haight
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