SOLUTION: Solve the following applications involving functions. 10. Business and finance. If the inventor in exercise 53 charges $4 per unit, then her profit for producing and selling x

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Question 122219: Solve the following applications involving functions.
10. Business and finance. 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 jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)

f%28x%29=2.25x-7000 Start with the given function.


f%282000%29=2.25%282000%29-7000 Plug in x=2000


f%282000%29=4500-7000 Multiply 2.25 and 2,000 to get 4,500


f%282000%29=-2500 Subtract 7,000 from 4,500 to get -2,500



So her profit is -2,500 dollars when she sells 2,000 units. In other words, she loses 2,500 dollars when she sells 2,000 units.







b)



f%28x%29=2.25x-7000 Start with the given function.


f%282000%29=2.25%285000%29-7000 Plug in x=5000


f%282000%29=11250-7000 Multiply 2.25 and 5,000 to get 11,250


f%282000%29=4250 Subtract 7,000 from 11,250 to get 4,250



So her profit is 4250 dollars when she sells 5,000 units







c)

What is the break-even point for sales? The break even point occurs when f%28x%29=0


f%28x%29=2.25x-7000 Start with the given function.


0=2.25x-7000 Replace f%28x%29 with 0


-2.25x=7000 Subtract 2.25x from both sides


x=-7000%2F-2.25 Divide both sides by -2.25


x=3111.11 Divide


So if you round to the nearest whole number, you get


x=3111


So the break-even point is when she sells about 3,111 units