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**Unformatted text preview: **r x
C (x) = 15000
80x + 150 = 15000
80x = 14850
14850
x=
= 185.625
80
Since we can only produce a whole number amount of PortaBoys, we can produce 185
PortaBoys for $15,000.
3. The restriction x ≥ 0 is the applied domain, as discussed in Section 1.5. In this context,
x represents the number of PortaBoys produced. It makes no sense to produce a negative
quantity of game systems.5
4. To ﬁnd C (0), we replace every occurrence of x with 0 in the formula for C (x) to get C (0) =
80(0) + 150 = 150. This means it costs $150 to produce 0 PortaBoys. The $150 is often called
the ﬁxed or start-up cost of this venture. (What might contribute to this cost?)
4 The similarity of this name to PortaJohn is deliberate.
Actually, it makes no sense to produce a fractional part of a game system, either, as we saw in the previous part
of this example. This absurdity, however, seems quite forgivable in some textbooks but not to us.
5 2.1 Linear Functions 119 5. If we were to graph y = C (x), we would be graphing the portion of the line y = 80x + 150
for x...

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