McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
8. Differences of Squares
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Exercise 47 Page 61

Practice makes perfect
a We are given an equation that models the sales of a particular CD.
S=- 25m^2+125mSince S is the number of CDs sold in thousands and m represents the number of months that the CD is on the market, in order to find in what month the sales are expected to stop, we will substitute 0 for S in the equation and solve for m.
S=- 25m^2+125m
0=- 25m^2+125m
0=- 25m(m-5)
We will apply the Zero Product Property to solve for m.
- 25m(m-5)=0
lc- 25m=0 & (I) m-5=0 & (II)
lm=0 m-5=0
lm_1=0 m_2=5
We ended with two values for m, 0 and 5. In the context of the situation, month 0 would be the month before sales started, so it makes sense that there would be no sales of the CD for that month. This leaves us with 5. This suggests that CD sales are expected to stop after 5 months.
b To find when the CD sales will peak, we will make a table for the x-values located between 0 and 5. We expect that S-values increase up to some point, then start to decrease. When the S-values start to decrease, we will stop.
m - 25m^2+125m S=- 25m^2+125m
0.5 - 25( 0.5)^2+125( 0.5) 56.25
1 - 25( 1)^2+125( 1) 100
2 - 25( 2)^2+125( 2) 150
3 - 25( 3)^2+125( 3) 150
4 - 25( 4)^2+125( 4) 100
5 - 25( 5)^2+125( 5) 0

We can see that the value of S is the same for months 2 and 3. This suggests that the maximum value of S, will be found halfway between these points. In other words, the peak CD sales are expected to hit about halfway through month 2, or month 2.5.

c To find the number of CDs sold at the peak, we will substitute 2.5 for m and solve for S.
S=- 25m^2+125m
S=- 25( 2.5)^2+125( 2.5)
S=- 25(6.25)^2+125(2.5)
S=- 156.25+312.5
S=156.25
Recall that S is calculated in units of thousands. Let's convert the value we found for S. 156.25 * 1000 =156 250 At its peak, 156 250 copies of the CD will be sold.