• anonymous
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?
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • chestercat
I got my questions answered at in under 10 minutes. Go to now for free help!
  • anonymous
Profit is \[(revenue) - (cost)\]Let P(q) be the profit as a function of q. Then,\[P(q)=R(q)-C(q)=600q-20000-3q^2\]The extrema within a domain, if they exist, will be found at those values q such that,\[P'(q)=0\]Doing this, we have\[P'(q)=600-6q:=0 \rightarrow q=100\]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),

Looking for something else?

Not the answer you are looking for? Search for more explanations.