Answer:
The production level will yield a maximum revenue when production level is approximately equal to 75,000.
Step-by-step explanation:
Revenue is given by
R(x) = [tex]-x^3 + 102x^2 + 1575x[/tex]
First, we differentiate R(x) with respect to x, to get,
[tex]\frac{d(R(x))}{dx} = -3x^2 + 204x + 1575[/tex]
Equating the first derivative to zero, we get,
[tex]\frac{d(R(x))}{dx} = 0\\\\ -3x^2 + 204x + 1575 = 0[/tex]
Solving, with the help of quadratic formula, we get,
[tex]x = 75, -7[/tex]
Again differentiation R(x), with respect to x, we get,
[tex]\frac{d^2(R(x))}{dx^2} = -6x + 204[/tex]
At x = 75
[tex]\frac{d^2(R(x))}{dx^2} < 0[/tex]
Thus, maxima occurs at x = 75 for R(x).
Thus, the production level will yield a maximum revenue when production level is approximately equal to 75,000.
