Suppose a baby food company has determined that its total revenue R for its food is given by R = - x 3 + 33 x 2 + 720 x where R is measured in dollars and x is the number of units (in thousands) produced. What production level will yield a maximum revenue?

Question

Shelby Pacheco

Mathematics

28 November, 13:59

Suppose a baby food company has determined that its total revenue R for its food is given by R = - x 3 + 33 x 2 + 720 x where R is measured in dollars and x is the number of units (in thousands) produced. What production level will yield a maximum revenue?

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Answers (1)

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Aryana Yates · Answer 1

A production level of 30 thousand units (x = 30)

Step-by-step 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