Question

use the following equation for the demand curve:
q=400−4p

(a) Enter the equation for revenue R as a function of price p :
R=

(b) What is the (instantaneous) rate of change of revenue with respect to price when the price is $ 13?

(c) At what price is the derivative of revenue with respect to price exactly zero?

Answer 1

@mathmale

Answer 2

Hope I can actually be of help; haven't worked with this type of algebra problem for some years now. But anyway: Revenue equals price times quantity.

If I sell you 50 pens for $2 each, the revenue coming my way would be (50 pens * $2.00/pen), or $100.

Thus if you have a formula relating price and quantity, which you do in q=400−4p, just solve this for p (price) and multiply the result by q (quantity). Read this carefully , please and let me know how much sense this makes to you before you get involved in calculations.

Answer 3

You want to express revenue in terms of price, p, which means eliminating q. Are you able to do that algebraically?

Answer 4

i'm not sure

Answer 5

I am contradicting myself, I'm afraid. We want to eliminate quantity (q), and keep price (p), so we must use q=400−4p; in other words, substitute 400-4p for quantity, q.

Again, Revenue is price times quantity; here it's R(p)=p(400-4p). Expand that, please.

Answer 6

I just got done doing this! I could probably help

Answer 7

okay R(p)=400p-4p^2

Answer 8

@kayders1997: Thank you for your offer, but at the moment Pulsified333 and I are doing fine.

Answer 9

I just say when you said I haven't done this in a long time sorry

Answer 10

saw

Answer 11

Next part: you want the instantaneous rate of change of R with respect to price. When you see "instantaneous rate of change," that should remind you of a sophisticated technique we should apply next. What is that technique?

Answer 12

is it taking the derivative?

Answer 13

Exactly. Please find dR/dp, the inst. rate of change of Revenue with respect to price.

Answer 14

R'(p)=400-8p

Answer 15

then i plug in 13 for P right?

Answer 16

That is the general formula for the inst. r. of c. The probelm statement gives you a certain value, $13, for price, p. Yes. plug in $13 and evalute the "inst. r. of c. of Rev. when p=$13."

Answer 17

R'(p)=400-8p, then R'(13)=400-8(13)=296

Answer 18

Great. But let's not forget the units of measurement. What are you measuring here?

Answer 19

$296?

Answer 20

Partially right. But "inst. rate of change" implies that TWO different quantities are changing, not just $. What's the other quantity? Hint: What is our independent variable in "inst. rate of change" here?

Answer 21

400

Answer 22

I don't think the question actual gives us units to work with

Answer 23

See if you can find your cue here: "What is the (instantaneous) rate of change of revenue with respect to price when the price"

Answer 24

@mathmale Revenue is the other quantity?

Answer 25

You're looking to find the inst. rate of change of Revenue (dependent variable) with respect to price.

Answer 26

oh

Answer 27

i am still confused on what the dependent variable is ;(

Answer 28

Revenue depends upon price.

Answer 29

so i was right?

Answer 30

Express your result, for the inst. rate of ch. of revenue with respect to price, when price is $13, as $296 per dollar increase in price.

Answer 31

oh okay

Answer 32

onto the last part? which i think i know how to do

Answer 33

0=400-8p, then solve for P which P=50

Answer 34

Great working with you. Let me know if you have further questions. I'm hitting the sack now, however. Very good. Be sure not to forget units of measurement.

Answer 35

I wont :D thank you

Answer 36

My pleasure! Good night.

Answer 37

Night