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Cost function is $c(q)=q$

Demand function is $q(p) = 100/p$ for $p\le 20$. And $q(p)=0$ for $p>20$.

In order to calculate profit maximizing output and price

$$max [p(q)*q-q]$$

And marginal revenue = marginal cost for FOC.

But I cannot get proper result. Why? Please help me to calculate this or give me a hint. Thanks.

  • Is $c(q)$ the total cost function or the unit cost function? – callculus42 Apr 28 '18 at 12:50
  • The question is not specified. I guess total cost function @callculus –  Apr 28 '18 at 17:00
  • In this case you have a constant revenue of 100 but increasing costs. Thus the maximum profit is where the cost are minimum.This is at q=0. Thus the maximum revenu=profit is $profit(0)=100$ – callculus42 Apr 28 '18 at 17:06
  • You said for $p>20$. So it is not possible for other one $p<20$. I’m confused a bit. Please expand your answer? @callculus –  Apr 28 '18 at 17:32
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    You´re right. Good catch. Then take $q=5\Rightarrow p=20$. The maximum profit is $95$ – callculus42 Apr 28 '18 at 17:38
  • Okay what you said is true. But there is a mistake for $p>20$. When $p>20$, demand $q=0$ so revenue = q(p)p= 0p=0. That for $p>20$, revenue is zero. Since price is greater than 20, profit becomes negative. @callculus –  Apr 28 '18 at 20:24
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    I don´t really understand what you mean. If p=20 the price is not greater that 20. It is true that if p>20 the profit is negative. This sounds logic to me since the revenue is 0. Above a price of 20 nobody buys a unit of the good. – callculus42 Apr 28 '18 at 20:47
  • @callculus I see. Thank you so much. Can you look at that question which is continue of above question. https://math.stackexchange.com/questions/2758044/profit-maximization-question –  Apr 28 '18 at 21:45
  • You´re welcome. I have already seen your question. I think I will look more closely tomorrow. Here in Germany it is almost midnight. – callculus42 Apr 28 '18 at 21:51
  • No worries whenever you have time. Then have a good night to you! And see you please @callculus –  Apr 28 '18 at 21:53

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