Optimization Help: Revenue Maximization

Calculus
Tutor: None Selected Time limit: 1 Day

Apr 17th, 2015

Revenue = cost per unit * total units 

x(500+3(90-x)= 500x+270x-3x^2 
770x-3x^2 
then take the derivative of this= 770-6x
770-6x=0 
x=128.3

Apr 17th, 2015

wrong one, this is the right one:

x(500+5(90-x)= 500x+450x-5x^2 
950x-5x^2=190x-x^2
then take the derivative of this= 190-2x
190-2x=0, x=95

so maximum revenue is 45125

Apr 17th, 2015

This answer did not come out correct

Apr 17th, 2015

revenue R = (90 - x)(500 + 5x) 

maximizing R is the same as maximizing R/10, call it r 

45000+450x-500x-5x^2=x^2-10x+4500

derivative=2x-10=0, x=5

so rent charged should be 500+5*5=525

my bad thought it's to miximize revenue

Apr 17th, 2015

hmm it is still not coming out correct

Apr 17th, 2015

revenue = R(x) = (500 + 5x)(90 - x) 
R(x) = -5x2 - 50x + 45000 
dR/dx = -10x-50=0, x=-5
500+5(-5)=475, how is it now? got the sign wrong before.........

Apr 17th, 2015

perfect thank you

Apr 17th, 2015

I'm sorry I had a lot of assignments yesterday and couldn't answer you in time. Hope you understand..I remember I still got wrong the other question. Can you invite me? I wanna try to fix that problem later :)

Apr 17th, 2015

do i just repost that question? it is still up and i don't want to say that I'm not satisfied with you lol...but I'm not sure how to invite you to it

Apr 17th, 2015

ok, then I will try to find it myself. Or you can repost as a free question. Your choice

Apr 17th, 2015

ok i reposted it

Apr 17th, 2015

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