Your boss has asked you to come up with a price model for one of the cell phone

Question

Your boss has asked you to come up with a price model for one of the cell phone models that your company produces. In the model, costs and length of service are directly related. You are also asked to examine the profit formula based on selling a certain number of the phones to determine profitability based on how many are sold.

1. Suppose a cell phone service costs $45 per month, along with a $60 activation fee.
1. Write a cost equation that relates the number of months of service and the total cost for that time of service. (Hint: Let t = the number of months of service and c = the total cost.)
2. Use the cost equation to determine the number of months of service for a cost of $1,185.
2. Your company has determined that the profit equation (in thousands of dollars) of producing x thousand smartphones is as follows:

Profit = –x2 + 110x – 1,000

(If the number of smartphones is 40,000, for example, then x = 40). The break-even point is the number of smartphones sold and produced that would result in a profit of zero.
Use your preferred method for solving the quadratic equation to determine the break-even point.

Explanation / Answer

1. First, cost must equal something. The two things that are available for cost are initial (one time) fee of 60, and second a monthly fee of 45. So
c= total fees
c= one time fee+ monthly fee times the number of months
c= 60 +45t

2.C= 1185 plug that value into previous equation
1185= 60+45t solve for t
1185=60+ 45t subtract 60 from both sides
-60 -60
1125=45t divide both sides by 45
/45 /45
25=T

3. The break even point is when P=0
so
0=-x^2+110x-1000 I like to have the x^2 to be positive, so divide everything by -1
/-1
0=x^2-110x+1000 (see how it changed the signs)
I like to get my quadratic into factored form if possible, so I start by taking 1000 and seeing if i can find factors that add up to -110
10 and 100 nope add up to 110 not negative
-20 and -50 nope add up to -70
-10 and -100 yes, add up to -110

now that i have factors of 1000 I can put together the factored form
0=(x-10)(x-100)
Now to check that i did it correctly i will Foil it back out
first x*x
outer x*-100
inner -10*x
last -10*-100
x^2-100x-10x+1000
it checks.
So now, if either x-10= 0 or x-100 equals zero the entire problem equals zero.
so lets take the first
x-10=0 add 10 to both sides
+10 +10
x=10

now the second
x-100=0 add 100 to both sides
+100 +100
x=100

now we have two answers, but which one is correct? The question askes for the break even point, which to me would mean the lowest number or the first number that would give us a 0. So, I would say the correct answer is that they break even when x=10.

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