Maximizing revenue: p = 10e^(-x), minimize R(x) = xp

[Linty Fresh]

A cosmetic company is planning the introduction and promotion of a new lipstick line. The marketing research department found that the demand in that city is given approximately by:

p = 10e^-x

where x thousand lipsticks were sold per week at a price of p dollars each.

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

OK, so . . .

R(x) = 10x*e^-x
R'(x) = uv' + vu' = (10x)(e^-x) + (e^-x)(10)
= 10*e^-x (x + 1)

And I get stuck here, because x can only equal -1, which is impossible. Now I'm assuming that if the derivative of e^x is e^x, then the derivative of e^-x should be e^-x. Is this where I'm going wrong?

 

[skeeter]

Re: Maximizing revenue

A cosmetic company is planning the introduction and promotion of a new lipstick line. The marketing research department found that the demand in that city is given approximately by:

p=10e^-x

where x thousand lipsticks were sold per week at a price of p dollars each.

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

OK, so . . .

R(x)=10x*e^-x
R'(x)=uv'+vu'=(10x)(-e^-x) + (e^-x)(10)
=10*e^-x (-x+1)
corrections in red

And I get stuck here, because x can only equal -1, which is impossible. Now I'm assuming that if the derivative of e^x is e^x, then the derivative of e^-x should be e^-x. the derivative of e^-x is -e^-x Is this where I'm going wrong?

 

[Linty Fresh]

Thanks for your help, skeeter. I have another question. Is the fact that the derivative of e^-x=-e^-x something that my textbook should have mentioned and I remembered, or should I have been able to work that out for myself?

 

[o_O]

If you knew that e^x is its own derivative and you know about the chain rule, then that's something you should figure out:

\(\displaystyle [f(g(x))]' = f'(g(x)) \cdot g'(x)\)

If we let f(x) = e^x and g(x) = -x, then:
f(g(x)) = e^g(x) = e^-x

Then, using the chain rule:
\(\displaystyle [f(g(x))]' = \frac{d}{dx} e^{g(x)} \cdot g'(x)\)

\(\displaystyle = e^{g(x)} \cdot (-x)'\)

\(\displaystyle = e^{-x} \cdot -1\)

So, (e^-x)' = -e^-x

 

[Linty Fresh]

Ah, gotcha! Unfortunately the chain rule is the next section after this one. That's what I get for buying my Calc textbook for a dollar at the library book sale!