Answer:
p = 59.11 dollars
Explanation:
Given
Price: p(x) = 8eˣ (0 ≤ x ≤ 2)
Revenue; R = x*p = 8xeˣ
p = ? when R be at maximum
We can apply
dR/dx = d(x*p)/dx = 0
⇒ d(8xeˣ)/dx = 8*(1*eˣ + x*eˣ) = 0
⇒ eˣ*(1 + x) = 0 ⇒ x = - 1
as x = - 1 ∉ [0, 2]
then, we have
p(0) = 8e⁰ = 8
R = 0*8 = 0
If x = 1
p(1) = 8e¹ ≈ 21.74
R = 1*21.74 = 21.74
If x = 2
p(2) = 8e² ≈ 59.11
R = 2*59.11 = 118.22
Implies that, R(x) is maximum at x = 2.
Thus, the price that maximize the revenue of the company is 59.11 dollars.
The price that maximizes the revenue is $23.85 per lipstick.
The relation between price and demand is:
p = 8e^-x
Using the relation:
ln(e^x) = x
We can isolate the demand to get:
x = -ln(p/8).
Then, the revenue is the product between the demand and the price, so the revenue is:
R = x*p = -ln(p/8)*p
We want to maximize this, to do it, we need to find the zeros of the derivate.
The derivate is:
R' = -ln(p/8) - p*(8/p) = ln(p/8) - 8
This is zero when:
ln(p/8) - 8 = 0
ln(p/8) = 8
If we use the exponential in both sides, we get:
exp(ln(p/8)) = exp(8)
p/8 = exp(8)
p = exp(8)*8 = 23,847.66
This would be the price of a thousand lipsticks (because x is in thousands).
So the price of each lipstick should be:
$23,847.66/1000 = $23.85.
The price that maximizes the revenue is $23.85 per lipstick.
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