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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
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p = -0.02(x - 30000p) so vertex is at x = 15000 so 15000 units will max the revenue
or using calculus max/min techniques p'=600-0.04x=0 x=15000
you only need to partially complete the square
whatever's easiest, I hate that trick.
p = -0.02(x^2 - 30000x)
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if they don't know calculus finding the derivative might be a problem
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.
I agree
I'm actually a calc student, though end of day its good to know both methods, thanks.
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