Question

Suppose that a tea company estimates that its monthly cost is C(c) = 500x² + 100x and its monthly revenue is R(2) = -0.6x3 + 700x2 – 400x + 300, where x is in thousands of boxes of tea sold. The profit is the difference between the revenue and the cost. What is the profit function, p(X)?
O A. P(x) = 0.6x3- 200x2 + 500x+ - 300

OB. P(x) = -0.6x3 + 1200x2 – 300x + 300

O C. P(x) = -0.6x3 + 200x2 – 500x + 300

D. P(xl = 0.6x3 + 200x2 – 500x + 300​

Answer 1

Answer:

p(x) = 0.6x³ - 200x² + 500x- 300

Step-by-step explanation:

Given the cost function and revenue function as:

C(x) = 500x² + 100x

R(x) = -0.6x³ + 700x² – 400x + 300

To get the profit function:

p(x) = C(x) - R(x)

p(x) = 500x² + 100x -(-0.6x³ + 700x² – 400x + 300)

open the parenthesis

p(x) = 500x² + 100x + 0.6x³ - 700x² + 400x - 300

p(x) = + 0.6x³+500x² - 700x² + 100x+ 400x- 300

p(x) = 0.6x³ - 200x² + 500x- 300

Hence the profit function is expressed as p(x) = 0.6x³ - 200x² + 500x- 300