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Mathematics 15 Online
OpenStudy (anonymous):

Find the number of units to be produced in order to maximize revenue if the demand function is given by p= 600x - 0.02x^2

OpenStudy (anonymous):

p = -0.02(x - 30000p) so vertex is at x = 15000 so 15000 units will max the revenue

OpenStudy (turingtest):

or using calculus max/min techniques p'=600-0.04x=0 x=15000

OpenStudy (anonymous):

you only need to partially complete the square

OpenStudy (turingtest):

whatever's easiest, I hate that trick.

OpenStudy (anonymous):

p = -0.02(x^2 - 30000x)

OpenStudy (anonymous):

if they don't know calculus finding the derivative might be a problem

OpenStudy (turingtest):

Understood; we're probably dealing with an algebra student I'm guessing. So your method would seem easier, but calc makes it much easier if you know it.

OpenStudy (anonymous):

I agree

OpenStudy (anonymous):

I'm actually a calc student, though end of day its good to know both methods, thanks.

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