SOLUTION: Solve the following applications involving functions. If the inventor in exercise 53 charges $4 per unit, then her profit for producing and selling x units is given by th

Algebra ->  Functions -> SOLUTION: Solve the following applications involving functions. If the inventor in exercise 53 charges $4 per unit, then her profit for producing and selling x units is given by th      Log On


   

Question 102423: Solve the following applications involving functions.



If the inventor in exercise 53 charges $4 per unit, then her
profit for producing and selling x units is given by the function
P(x)=2.25x-7000

(a) What is her profit if she sells 2000 units?
(b) What is her profit if she sells 5000 units?
(c) What is the break-even point for sales?
Answer by JP(22) About Me  (Show Source):
You can put this solution on YOUR website!
If the function is P(x)=2.25x-7000
A.) For every instance of x, you must plug in 2000 units since x is equal to the units sold. Therefore, P(2000)=2.25(2000)-7000
2.25(2000)=4500-7000=-$2500 profit.
B.)For every instance of x, you must plug in 5000 units since x is equal to the units sold. Therefore, p(5000)=2.25(5000)-7000
2.25(5000)=11250, then subtract 7000 to get a profit of $4250
C.)To get the break-even point you must set the units to x.
Your equation is P(x)=2.25(x)-7000=0
2.25(x)=2.25x-7000=0 Then add 7000 to the other side.
2.25x=7000 divide now to get the x variable alone
x=7000/2.25
x=$3111.11
Plug it into the original function to verify your work.
P(3111.11)=2.25(3111.11)-7000
2.25(3111.11)=7000 then subtract 7000
7000-7000=0 or your break-even point
I hope this helps...however, please verify that the function p(x)=2.25x-7000 is the correct function given by your book. If you have any further questions let me know.