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

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

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

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

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

p was 5/3

i have all my work attached

i tried putting that in but it doesnt work:/

So it should be 0.

i also tried zero.

The rate at which 50 is changing is 0.

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

i tried 0 50 and it's wrong:/

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

i have it up there @pgpilot326

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

i mean, is the work right or wrong?

then is part A? wrong?

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

how is dx/dt= -9?

where did the 100 come from?

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

in the pic...

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

x is 100.

not 30....

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

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

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

dx/dt, sorry

so wouldn't it be -30?

dx/dt

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

yes

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

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

dp/dt is $0.5/week

i tried. -30 as dx/dt

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

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

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

i kept trying everything, nothing works.

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

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

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

revenue = price x quantity sold

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