Question

help: related rate problem with demand: (attached)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

p was 5/3

Answer 7

i have all my work attached

Answer 8

@pgpilot326

Answer 9

revenue = price*quantity

R = p*x
but px = 50 by definition of demand function

so revenue is constant at 50

Answer 10

i tried putting that in but it doesnt work:/

Answer 11

So it should be 0.

Answer 12

i also tried zero.

Answer 13

The rate at which 50 is changing is 0.

Answer 14

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

Answer 15

i tried 0 50 and it's wrong:/

Answer 16

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

Answer 17

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

Answer 18

\[p \frac{ dx }{ dw }+x \frac{ dp }{ dw }=0\]. you can find the price based on the demand/price equation
\[xp=50\]
plug in what you know to find what you need.

Answer 19

i have it up there @pgpilot326

Answer 20

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

Answer 21

i mean, is the work right or wrong?

Answer 22

then is part A? wrong?

Answer 23

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

Answer 24

@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

Answer 25

how is dx/dt= -9?

Answer 26

where did the 100 come from?

Answer 27

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

Answer 28

@dumbcow it's okay

Answer 29

in the pic...

Answer 30

?

Answer 31

@dumbcow im still not sure how -9 came from

Answer 32

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

Answer 33

x is 100.

Answer 34

not 30....

Answer 35

@pgpilot326

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

dx/dt, sorry

Answer 40

so wouldn't it be -30?

Answer 41

dx/dt

Answer 42

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

Answer 43

yes

Answer 44

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

Answer 45

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

Answer 46

dp/dt is $0.5/week

Answer 47

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

Answer 48

i tried. -30 as dx/dt

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

i kept trying everything, nothing works.

Answer 53

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

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

Answer 54

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...

Answer 55

that gives you same thing....

dx/dt = -9

Answer 56

as far as revenue, not sure what to do

the rate has to be 0

Answer 57

revenue = price x quantity sold

Answer 58

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.

Answer 59

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