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anonymous
 5 years ago
Revenue is given by R(q)=600q and cost is given by C(q)= 20000+3q^2
At what quantity, q, is the profit maximized?
anonymous
 5 years ago
Revenue is given by R(q)=600q and cost is given by C(q)= 20000+3q^2 At what quantity, q, is the profit maximized?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Profit is \[(revenue)  (cost)\]Let P(q) be the profit as a function of q. Then,\[P(q)=R(q)C(q)=600q200003q^2\]The extrema within a domain, if they exist, will be found at those values q such that,\[P'(q)=0\]Doing this, we have\[P'(q)=6006q:=0 \rightarrow q=100\]The fact that the profit function is quadratic with negative coefficient on x^2 is enough to show that this is a parabola which is concave down. The extremum you find will therefore be maximal. Otherwise (in other situations where the geometry isn't clear, take the second derivative and test the point you found. If the second derivative at that point turns out negative, you have a maximum at that point. If it's positive, you have a minimum),
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