Answer:
A production level of 30 thousand units (x = 30)
Explanation:
To find the production level (value of x) that will yield the maximum revenue, we can take the derivative of the function R in relation to x and find when it is equal to 0:
dR/dx = -3x2 + 66x + 720 = 0
x2 - 22x - 240 = 0
Solving the quadratic equation using Bhaskara's formula, we have:
Delta = (-22)^2 + 4*240 = 1444
sqrt(Delta) = 38
x1 = (22 + 38)/2 = 30
x2 = (22 - 38)/2 = -8
The negative value is not valid for our problem, so we have that the value that gives the maximum revenue is x = 30