A baby food manufacturer finds that its revenue for selling x units (in thousands) is R(x) = - x3+63x2+12000x

Question

A baby food manufacturer finds that its revenue for selling x units (in thousands) is

R(x) = - x³+63x²+12000x

What production level maximizes revenue?

Answer 1

Let's find the extremums of R:

R'(x)= -3x² + 126x + 12000 = 0

x² - 42x- 4000 =0

x_{1, 2} = 21 ± √4441

Solving the quadratic equation, we get x₁ ≈ -45.64 and x₂ ≈ 87.64.

x₁ - cannot be a solution because it is negative.

Now, we need to understand what kind of extremum we found. Let's take the second derivative:

R''(x) = -6x + 126

R''(87.64) = -399.84 < 0 -> x ≈ 87.64 is a maximum point.

Answer: x ≈ 87641 units

Answer 2

Another option is to find the derivative, graph it.

Discard the negative solution, because the manufacturer can't make a negative number of units.

The positive x intercept of the derivative -3x^{2}+126x+12000 is 87.64083...

The question states it's in thousands, so it's 87,641 units.