The price per unit is given by the equation p=18-2ln(x). What is the marginal revenue at x=400? And what should the number of units sold be in order to maximize revenue?
To find the marginal revenue at x = 400, we need to find the derivative of the revenue function with respect to x.
The revenue function is given by R(x) = xp(x), where p(x) is the price function.
First, let's differentiate the price function, p(x) = 18 - 2ln(x), with respect to x:
dp(x)/dx = -2(1/x) = -2/x
Next, we can calculate R(x) by substituting p(x) into the revenue function:
R(x) = x(18 - 2ln(x)) = 18x - 2xln(x)
Now, let's differentiate R(x) with respect to x:
dR(x)/dx = 18 - 2ln(x) - 2x(1/x) = 18 - 2ln(x) - 2
To find the marginal revenue, we substitute x = 400 into dR(x)/dx:
Marginal revenue at x = 400 = 18 - 2ln(400) - 2
Now, let's calculate the marginal revenue:
Marginal revenue = 18 - 2ln(400) - 2
= 18 - 2(5.991) - 2
= 18 - 11.982 - 2
= 4.018
So, the marginal revenue at x = 400 is approximately $4.018.
To determine the number of units sold (x) at which the revenue is maximized, we need to find the critical points of the revenue function. This occurs when dR(x)/dx = 0.
18 - 2ln(x) - 2 = 0
Simplifying the equation:
-2ln(x) = -16
Dividing by -2:
ln(x) = 8
Taking the exponential:
e^(ln(x)) = e^8
x = e^8
Therefore, to maximize revenue, the number of units sold (x) should be approximately e^8 units.