MATH SOLVE

4 months ago

Q:
# 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^2-200B) C(x)= 60x^2+200C) C(x)=2x+60x^2+200D) C(x)= -2x-60x^2-200

Accepted Solution

A:

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

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