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