help: related rate problem with demand: (attached)

- anonymous

help: related rate problem with demand: (attached)

- jamiebookeater

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- anonymous

##### 1 Attachment

- anonymous

i understand how to find dx/dt but im not so sure about part b) to find the revenue

- anonymous

so my answer to a) idx/dt=100.

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- anonymous

##### 1 Attachment

- e.mccormick

Demand is how many are sold. Revenue is income from the sales.

- anonymous

yes i know, but how does i figure it out for part b

- e.mccormick

So you solved for p and took the first derivative of that?

- anonymous

p was 5/3

- anonymous

i have all my work attached

- anonymous

@pgpilot326

- dumbcow

revenue = price*quantity
R = p*x
but px = 50 by definition of demand function
so revenue is constant at 50

- anonymous

i tried putting that in but it doesnt work:/

- e.mccormick

So it should be 0.

- anonymous

i also tried zero.

- e.mccormick

The rate at which 50 is changing is 0.

- e.mccormick

Hmm... very odd. If it is not the derivative or the constant and they gave you nothing else....

- anonymous

i tried 0 50 and it's wrong:/

- dumbcow

how did you get dx/dt = 100 ?
if price is increasing , then quantity should be decreasing

- e.mccormick

I wonder if they mean Marginal Revenue:
http://economics.about.com/od/production/ss/Marginal-Revenue-And-The-Demand-Curve.htm

- anonymous

##### 1 Attachment

- 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

i have it up there @pgpilot326

- anonymous

it seems as though the revenue is unchanged, holding constant at $50

- anonymous

i mean, is the work right or wrong?

- anonymous

then is part A? wrong?

- anonymous

as such the rate of change of revenue is $0.

- 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

how is dx/dt= -9?

- anonymous

where did the 100 come from?

- dumbcow

solving the equation you had
30(.5) + (5/3)dx/dt = 0

- anonymous

@dumbcow it's okay

- anonymous

in the pic...

- anonymous

?

- anonymous

@dumbcow im still not sure how -9 came from

- anonymous

x=30, p = 5/3, dp/dw = .5 and solve for dx/dw, yeah?

- anonymous

x is 100.

- anonymous

not 30....

- anonymous

@pgpilot326

- anonymous

what? where did that come from? it says "when demand is 30"

- anonymous

and why can't we plug in dx/dt as 30 if dp/dt is 0.5

- anonymous

gx/dt is the change in demand relative to th echange in time

- anonymous

dx/dt, sorry

- anonymous

so wouldn't it be -30?

- anonymous

dx/dt

- dumbcow

hmm, ok to get price of 5/3 you plugged in x=30 right

- anonymous

yes

- dumbcow

so why would x change its value....if its 30 its 30 :)

- anonymous

so bc price is 0.5 to 1.67.... then we have a change?

- anonymous

dp/dt is $0.5/week

- 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

i tried. -30 as dx/dt

- anonymous

i tried 0 and 50 for part b still wrong ::??

- dumbcow

in other words, you would never say the price is 0.5

- anonymous

okay can i see the answers to check if theyre right?

- anonymous

i kept trying everything, nothing works.

- anonymous

(a) The rate at which the demand is changing is .
(b) The rate at which the revenue is changing is .

- 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

that gives you same thing....
dx/dt = -9

- dumbcow

as far as revenue, not sure what to do
the rate has to be 0

- anonymous

revenue = price x quantity sold

- 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

http://en.wikipedia.org/wiki/Marginal_revenue

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