A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
mathcalculus
 one year ago
help: related rate problem with demand: (attached)
mathcalculus
 one year ago
help: related rate problem with demand: (attached)

This Question is Closed

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i understand how to find dx/dt but im not so sure about part b) to find the revenue

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so my answer to a) idx/dt=100.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Demand is how many are sold. Revenue is income from the sales.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0yes i know, but how does i figure it out for part b

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0So you solved for p and took the first derivative of that?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i have all my work attached

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0revenue = price*quantity R = p*x but px = 50 by definition of demand function so revenue is constant at 50

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i tried putting that in but it doesnt work:/

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0So it should be 0.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i also tried zero.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0The rate at which 50 is changing is 0.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0Hmm... very odd. If it is not the derivative or the constant and they gave you nothing else....

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i tried 0 50 and it's wrong:/

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0how did you get dx/dt = 100 ? if price is increasing , then quantity should be decreasing

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.0I wonder if they mean Marginal Revenue: http://economics.about.com/od/production/ss/MarginalRevenueAndTheDemandCurve.htm

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.1\[p \frac{ dx }{ dw }+x \frac{ dp }{ dw }=0\]. you can find the price based on the demand/price equation \[xp=50\] plug in what you know to find what you need.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i have it up there @pgpilot326

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.1it seems as though the revenue is unchanged, holding constant at $50

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i mean, is the work right or wrong?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0then is part A? wrong?

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.1as such the rate of change of revenue is $0.

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0@mathcalculus , sorry didnt see you posted your work ok your work is good but seems you just plugged in wrong numbers x = 30 dp/dt = 0.5 p = 5/3 dx/dt = 9

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0how is dx/dt= 9?

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.1where did the 100 come from?

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0solving the equation you had 30(.5) + (5/3)dx/dt = 0

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0@dumbcow it's okay

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0@dumbcow im still not sure how 9 came from

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.1x=30, p = 5/3, dp/dw = .5 and solve for dx/dw, yeah?

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.1what? where did that come from? it says "when demand is 30"

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0and why can't we plug in dx/dt as 30 if dp/dt is 0.5

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.1gx/dt is the change in demand relative to th echange in time

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so wouldn't it be 30?

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0hmm, ok to get price of 5/3 you plugged in x=30 right

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0so why would x change its value....if its 30 its 30 :)

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so bc price is 0.5 to 1.67.... then we have a change?

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0oh i see, there is a difference between current price and rate price is changing at this moment, x=30 and p=5/3 the rate they are changing is dp/dt = 0.5 , dx/dt = 9 so a week later the price will increase and quantity will decrease

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i tried. 30 as dx/dt

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i tried 0 and 50 for part b still wrong ::??

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0in other words, you would never say the price is 0.5

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0okay can i see the answers to check if theyre right?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i kept trying everything, nothing works.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0(a) The rate at which the demand is changing is . (b) The rate at which the revenue is changing is .

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.1the only thing I can think of is to write x as a function of p, i.e., \[x=\frac{ 50 }{ p }\] Then \[\frac{ dx }{ dt }=\frac{ 50 }{ p ^{2} }\times \frac{ dp }{ dt }\] See what that gets you...

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0that gives you same thing.... dx/dt = 9

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0as far as revenue, not sure what to do the rate has to be 0

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.1revenue = price x quantity sold

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.1Marginal revenue is equal to the ratio of the change in revenue for some change in quantity sold to that change in quantity sold. This can also be represented as a derivative when the change in quantity sold becomes arbitrarily small. More formally, define the revenue function to be the following \[ R(q)=P(q)\cdot q \]. By the product rule, marginal revenue is then given by \[ R'(q)=P(q) + P'(q)\cdot q\]. For a firm facing perfect competition, price does not change with quantity sold (P'(q)=0), so marginal revenue is equal to price.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.