Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

help: related rate problem with demand: (attached)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

1 Attachment
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
revenue = price*quantity R = p*x but px = 50 by definition of demand function so revenue is constant at 50
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 wonder if they mean Marginal Revenue: http://economics.about.com/od/production/ss/Marginal-Revenue-And-The-Demand-Curve.htm
1 Attachment
\[p \frac{ dx }{ dw }+x \frac{ dp }{ dw }=0\]. you can find the price based on the demand/price equation \[xp=50\] plug in what you know to find what you need.
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.
@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
how is dx/dt= -9?
where did the 100 come from?
solving the equation you had 30(.5) + (5/3)dx/dt = 0
@dumbcow it's okay
in the pic...
?
@dumbcow im still not sure how -9 came from
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
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
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 .
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...
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
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.
http://en.wikipedia.org/wiki/Marginal_revenue

Not the answer you are looking for?

Search for more explanations.

Ask your own question