The revenue, in thousands of dollars, that a company earns selling lawnmowers can be modeled by R (x) = 90x-x^2 and the company's total profit, in thousands of dollars, after selling x lawnmowers can be modeled by p (x) = - x^2+30x-200. Which function represents the company's cost, in thousands of dollars, for producing lawnmowers? (Recall that profit equals revenue minus cost.)

A) C (x) = - 60x^2-200

B) C (x) = 60x^2+200

C) C (x) = 2x+60x^2+200

D) C (x) = - 2x-60x^2-200

Question

Marina Dorsey

Mathematics

5 October, 04:31

The revenue, in thousands of dollars, that a company earns selling lawnmowers can be modeled by R (x) = 90x-x^2 and the company's total profit, in thousands of dollars, after selling x lawnmowers can be modeled by p (x) = - x^2+30x-200. Which function represents the company's cost, in thousands of dollars, for producing lawnmowers? (Recall that profit equals revenue minus cost.)

A) C (x) = - 60x^2-200

B) C (x) = 60x^2+200

C) C (x) = 2x+60x^2+200

D) C (x) = - 2x-60x^2-200

+4

Answers (1)

Know the Answer?

Shyla Hunter · Answer 1

Profit, P is given by:

P=R-C

where

R is the revenue

C is the cost

this implies that:

C=R-P

from the information given, the revenue and the cost functions are given by:

R (x) = 90x²-x

P (x) = 30x²-x-200

thus the cost function will be:

C (x) = R (x) - P (x)

substituting our values we shall have:

C (x) = 90x²-x - (30x²-x-200)

C (x) = (90x²-30x²-x+x+200)

C (x) = 60x²+200

The answer is B