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mathcalculus
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help: related rate problem with demand: (attached)
 11 months ago
 11 months ago
mathcalculus Group Title
help: related rate problem with demand: (attached)
 11 months ago
 11 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
so my answer to a) idx/dt=100.
 11 months ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
Demand is how many are sold. Revenue is income from the sales.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yes i know, but how does i figure it out for part b
 11 months ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
So you solved for p and took the first derivative of that?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
p was 5/3
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i have all my work attached
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@pgpilot326
 11 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i tried putting that in but it doesnt work:/
 11 months ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
So it should be 0.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i also tried zero.
 11 months ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
The rate at which 50 is changing is 0.
 11 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i tried 0 50 and it's wrong:/
 11 months ago

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

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

pgpilot326 Group TitleBest 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.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i have it up there @pgpilot326
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
it seems as though the revenue is unchanged, holding constant at $50
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i mean, is the work right or wrong?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
then is part A? wrong?
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
as such the rate of change of revenue is $0.
 11 months ago

dumbcow Group TitleBest 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
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
how is dx/dt= 9?
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
where did the 100 come from?
 11 months ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
solving the equation you had 30(.5) + (5/3)dx/dt = 0
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@dumbcow it's okay
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
in the pic...
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@dumbcow im still not sure how 9 came from
 11 months ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
x is 100.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
not 30....
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@pgpilot326
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
what? where did that come from? it says "when demand is 30"
 11 months ago

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

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
gx/dt is the change in demand relative to th echange in time
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
dx/dt, sorry
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
so wouldn't it be 30?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
dx/dt
 11 months ago

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

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
so why would x change its value....if its 30 its 30 :)
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
so bc price is 0.5 to 1.67.... then we have a change?
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
dp/dt is $0.5/week
 11 months ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
oh 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
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i tried. 30 as dx/dt
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i tried 0 and 50 for part b still wrong ::??
 11 months ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
in other words, you would never say the price is 0.5
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
okay can i see the answers to check if theyre right?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i kept trying everything, nothing works.
 11 months ago

mathcalculus Group TitleBest 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 .
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
the 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...
 11 months ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
that gives you same thing.... dx/dt = 9
 11 months ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
as far as revenue, not sure what to do the rate has to be 0
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
revenue = price x quantity sold
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
Marginal 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.
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.1
http://en.wikipedia.org/wiki/Marginal_revenue
 11 months ago
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