Exponential Functions.......... I'M STUCK....... ++++++++++++?

Question

Here is the problem and what I did so far.
I get stuck at a certain point.
I would really appreciate some help.


Demand Equation: p = 12e^-.001x
0 ≤ x ≤ 2

A) Find the rate of change of price with respect to demand when the weekly demand is 800 lipsticks and interpret.

B) At what price will the weekly revenue R(x) = xp be maximum? What is the maximum weekly revenue in the test city?

C) Graph R for 0 ≤ x ≤ 2.

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This is what I have so far:

p' = -0.012e^(-.001x)
p' = -0.012e^(-.001 * 800)
p' = -.005

R(x) = x*p
R(x) = x (12e^.001x)
R(x) = 12xe^.001x

R'(x) = 12xe^(.001x) + 12e^(.001x)

Okay... How in the world am I supposed to get x from THAT?
Because once I get x, I plug it back into the revenue function to get the maximum revenue and plug it into the demand function.

I just need to know how to get "X" from that.
And did I do it right so far?

Answer 1

ok up to
R'(x) = 12xe^(.001x) + 12e^(.001x)
which should be
R'(x) = -0.012xe^(-.001x) + 12e^(-.001x)
= 12e^(-0.001x) (1-0.001x)
i.e. max at x = 1000

Answer 2

R(x) = 12 x*e-.001x (you forgot the -)
so R´(x) =12 ( e^-0.001x - 0.001x e^0.001x)=12e^0.001x(1-0.001x)
As the exponential is always positive
R´(x)= 0
1-0.001x=0 so x= 1,000

Answer 3

I really wanted to help but this was my worst subject in school and I'm amazed you got this far! Seriously, if your not going to be too late get help from the math club. PEACE.

Vin