I need help with this problem

[chica2006]

the weekly demand for an apartment in a city is p=1200 - .5x, where p denotes the monthly rent for the apartment and x denotes the number of apartments rented. monthly cost associated with renting a total of x apartements is C(x) = 30,000 + 400x.

(a) show that the revenue function R(x) is 1200x - .5x²

(b) find the profit function P(x)

(c) find the marginal profit function P'(x)

(d) at what price is the profit a maximum?

hey ... I really need help with this problem ... thanks
===========================================
THis is what i have so far. Is this right???

a)R(x) = x(1200 - .5x)
=x(1200-.5x)
R(x)=1200x-.5x²

b) R(x)=R(x)-C(x)
=(1200x - .5x² ) -(30,000 + 400x)
P(x)= -.5x² +800x-30,000

c)P'(x) = -x+800

d)p'(x) = -x+800
-x=-800
p"(x)= -1 --> maximum

p(800) = -.5(800)² +800(800) - 30,000
=$29,000

price per unit for max profit is: p=1200-.5(800)
= $800

 

[Gene]

a) is good (of course)
b) is good except for a minor typo.

R(x)=R(x)-C(x) is
P(x)=R(x)-C(x)

c) is good but should be written
x=800

d) the equation is good, but you lost a zero.
P(800) = -.5(800)² +800(800) - 30,000
=$290,000

You should not mix P and p. They are two different variables in most equations.

All in all a good job.