anonymous
  • anonymous
help: related rate problem with demand: (attached)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
i understand how to find dx/dt but im not so sure about part b) to find the revenue
anonymous
  • anonymous
so my answer to a) idx/dt=100.

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anonymous
  • anonymous
1 Attachment
e.mccormick
  • e.mccormick
Demand is how many are sold. Revenue is income from the sales.
anonymous
  • anonymous
yes i know, but how does i figure it out for part b
e.mccormick
  • e.mccormick
So you solved for p and took the first derivative of that?
anonymous
  • anonymous
p was 5/3
anonymous
  • anonymous
i have all my work attached
anonymous
  • anonymous
@pgpilot326
dumbcow
  • dumbcow
revenue = price*quantity R = p*x but px = 50 by definition of demand function so revenue is constant at 50
anonymous
  • anonymous
i tried putting that in but it doesnt work:/
e.mccormick
  • e.mccormick
So it should be 0.
anonymous
  • anonymous
i also tried zero.
e.mccormick
  • e.mccormick
The rate at which 50 is changing is 0.
e.mccormick
  • e.mccormick
Hmm... very odd. If it is not the derivative or the constant and they gave you nothing else....
anonymous
  • anonymous
i tried 0 50 and it's wrong:/
dumbcow
  • dumbcow
how did you get dx/dt = 100 ? if price is increasing , then quantity should be decreasing
e.mccormick
  • e.mccormick
I wonder if they mean Marginal Revenue: http://economics.about.com/od/production/ss/Marginal-Revenue-And-The-Demand-Curve.htm
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
\[p \frac{ dx }{ dw }+x \frac{ dp }{ dw }=0\]. you can find the price based on the demand/price equation \[xp=50\] plug in what you know to find what you need.
anonymous
  • anonymous
i have it up there @pgpilot326
anonymous
  • anonymous
it seems as though the revenue is unchanged, holding constant at $50
anonymous
  • anonymous
i mean, is the work right or wrong?
anonymous
  • anonymous
then is part A? wrong?
anonymous
  • anonymous
as such the rate of change of revenue is $0.
dumbcow
  • dumbcow
@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
anonymous
  • anonymous
how is dx/dt= -9?
anonymous
  • anonymous
where did the 100 come from?
dumbcow
  • dumbcow
solving the equation you had 30(.5) + (5/3)dx/dt = 0
anonymous
  • anonymous
@dumbcow it's okay
anonymous
  • anonymous
in the pic...
anonymous
  • anonymous
?
anonymous
  • anonymous
@dumbcow im still not sure how -9 came from
anonymous
  • anonymous
x=30, p = 5/3, dp/dw = .5 and solve for dx/dw, yeah?
anonymous
  • anonymous
x is 100.
anonymous
  • anonymous
not 30....
anonymous
  • anonymous
@pgpilot326
anonymous
  • anonymous
what? where did that come from? it says "when demand is 30"
anonymous
  • anonymous
and why can't we plug in dx/dt as 30 if dp/dt is 0.5
anonymous
  • anonymous
gx/dt is the change in demand relative to th echange in time
anonymous
  • anonymous
dx/dt, sorry
anonymous
  • anonymous
so wouldn't it be -30?
anonymous
  • anonymous
dx/dt
dumbcow
  • dumbcow
hmm, ok to get price of 5/3 you plugged in x=30 right
anonymous
  • anonymous
yes
dumbcow
  • dumbcow
so why would x change its value....if its 30 its 30 :)
anonymous
  • anonymous
so bc price is 0.5 to 1.67.... then we have a change?
anonymous
  • anonymous
dp/dt is $0.5/week
dumbcow
  • dumbcow
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
anonymous
  • anonymous
i tried. -30 as dx/dt
anonymous
  • anonymous
i tried 0 and 50 for part b still wrong ::??
dumbcow
  • dumbcow
in other words, you would never say the price is 0.5
anonymous
  • anonymous
okay can i see the answers to check if theyre right?
anonymous
  • anonymous
i kept trying everything, nothing works.
anonymous
  • anonymous
(a) The rate at which the demand is changing is . (b) The rate at which the revenue is changing is .
anonymous
  • anonymous
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...
dumbcow
  • dumbcow
that gives you same thing.... dx/dt = -9
dumbcow
  • dumbcow
as far as revenue, not sure what to do the rate has to be 0
anonymous
  • anonymous
revenue = price x quantity sold
anonymous
  • anonymous
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.
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Marginal_revenue

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