SOLUTION: 21. (See Week 4 Lecture page “Solving Rational Inequalities” for a starting point.) Suppose that a factory’s cost (in dollars) for making x cars is C(x) = 7020x + 8460 Then t

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 21. (See Week 4 Lecture page “Solving Rational Inequalities” for a starting point.) Suppose that a factory’s cost (in dollars) for making x cars is C(x) = 7020x + 8460 Then t      Log On


   



Question 1089704: 21. (See Week 4 Lecture page “Solving Rational Inequalities” for a starting point.) Suppose that a factory’s cost (in dollars) for making x cars is
C(x) = 7020x + 8460
Then the average cost per car is
𝐶′(𝑥)= 7020𝑥+8460 /𝑥
How many cars must the factory make each day to ensure the average cost per car is no more than $7200?

Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.
From the condition, this phrase "the average cost per car is no more than $7200"

means that an inequality takes place


%287020x+%2B+8460%29%2Fx <= 7200.    (1)


And they want you solve this inequality for x.  


To solve it, multiply both sides by x.  Since x is positive, by its meaning, you can do it without problems. You will get


7020x + 8460 <= 7200x.


Simplify it by subtracting the number 7020 from both sides. You will get

8640 <= 7200x - 7020,   

8640 <= 180x,

x >= 8640%2F180 = 48.


So,  x >= 48  is your solution and is your answer.

Solved.