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Calculus1 13 Online
OpenStudy (anonymous):

The manager of a large apartment complex knows from experience that 110 units will be occupied if the rent is 324 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 6 dollar increase in rent. Similarly, one additional unit will be occupied for each 6 dollar decrease in rent. What rent should the manager charge to maximize revenue?

OpenStudy (dumbcow):

Revenue = rent*units rent = 324 + 6x units = 110 - x \[R(x) = (324+6x)(110-x) = 35640+336x-6x^{2}\] to maximize revenue , set derivative equal to zero \[R'(x) = 336-12x = 0\] \[x = 336/12 = 28\] max rent = 324+6*28 = 492

OpenStudy (anonymous):

Thank you for your help!

OpenStudy (dumbcow):

yw

OpenStudy (whpalmer4):

@mapleryan If you recognize that \(R(x)\) is a parabola of the form \(ax^2+bx+c\), you can find the x-value of the vertex (point of maximum revenue, in this case) by \(x = -b/2a\). Actually the same calculation as with the derivative, but without the need to know the care and feeding of derivatives if you haven't advanced that far yet.

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