Question

Suppose that the demand of a certain item is Q=60e^(−0.02p). Remember, elasticity is given by the function: E=(dQ/dp)p/Q.
Evaluate the elasticity at p=50

I cant seem to get the right answer for this. Please show all steps so i can see where I went wrong

Answer 1

what was the derivative of Q that you got?

Answer 2

-2e^(-.02p)
Sorry for the late response, computer problems

Answer 3

I'm not confident that I got my Q right though

Answer 4

ummm isn't \[\frac{dQ}{dp} = -0.02\times 60e^{-0.02p}\]

Answer 5

what rule did you use to determine that?

Answer 6

its a chain rule problem

when differentiating exponentials\[y = e^{f(x)}.... then ... \frac{dy}{dx} = f'(x) \times e^{f(x)}\]

Answer 7

so it wouldn't be -0.02P (e^(-0.02p-1)) x (-0.02p)' ? is that not the chain rule?

Answer 8

we are differentiating with respect to p

so f(p) = -0.02p

so f'(p) = -0.02

Answer 9

and it that because when you apply the product rule to that: -0.02 (1) + (0)p?
I'm having a difficult time understanding this section, so sorry in advance for all the questions

Answer 10

well that the basic derivative rule its like f(x) = 10x f'(x) = 10

Answer 11

ok i think i'm starting to understand it

Answer 12

so the derivative is \[\frac{dQ}{dp} = - 1.2 e^{-0.02p}\]

Answer 13

so looking at your elasticity equation you will have

\[E = \frac{-1.2e^{-0.02p} \times p}{60e^{-0.02p}}\]

Answer 14

right

Answer 15

the last set would be to substitute p = 50 and evaluate

Answer 16

I think E = -1

Answer 17

but thats a rough guess

Answer 18

you would be correct

Answer 19

can I ask your help with a question of the same form to ensure i'm doing it right?

Answer 20

well I cheated and cancelled the e^{-0.02p}

so its (-1.2 x 50)/60 = -60/60 = -1

Answer 21

ahh ok

Answer 22

if Q(p)=a−bp where 0≤p≤ab.
would Q'(p) be 1-b+p? or am I still doing that wrong?

Answer 23

well it would appear that you are differnentiating with respect to p...

so then a is a constant and the derivative of the constant is 0

for me

Q'(p) = -b

Answer 24

darn it, why can't I figure this stuff out? Can you explain why it would be -b? sorry!

Answer 25

well p is the independent variable in the equation and Q depends on p and is called the dependent variable.

so you will need to differentiate Q with respect to p....

so identify terms with a p... and then apply the differentiate. other terms become 0

Answer 26

so everything that doesnt have a p becomes zero, and anything attatched to the p or independent variable is differentiated? is that right?

Answer 27

thats correct... because using the index rule

\[f(x) = x^n..... f'(x) = nx^{n-1}\]

so if you have

\[f(x) = 3x ... or ...f(x) = 3x^1....f'(x) = 3x^{1 -1} ....or....f'(x) = 3x^0 = 3\]

Answer 28

ok so the -b was kind of like -b(1) P^(1-1)?

Answer 29

thats right.. anything to the power zero = 1

Answer 30

oook. So if E(p)= Q'(p) (p/Q(p)) I would take -b x p divided by a-bp ? would that be correct?

Answer 31

yep... that seems to make sense...

Answer 32

so what do I do when they ask for d^2E/dp^2. What does that mean?

Answer 33

I'll assume its the 2nd derivative.

find the 1st derivative of E with respect to p

when you have that neat and tidy differentiate agian with respect to p

Answer 34

using your earlier information about E(p) you'' need the quotient rule

Answer 35

would that then be (a-bp)(1)-(-bp)(-b) divded by (a-bp)^2??

Answer 36

I haven't do it... getting a bit tired... sorry

Answer 37

thank you for all your help, i greatly appreciate it! :)

Answer 38

good luck