Question

The function  n(t)=−4t2+100t represents the number of automobile parts produced at a factory after t hours. The function c(n)=20,000+5,000n  represents the cost in dollars of producing n car parts. What is  (c∘n)(t) and what does it mean? What is the domain?

Answer 1

To find \((c \circ n)(t)\), we need to substitute the function \(n(t)\) into the function \(c(n)\).

1. **Substituting \(n(t)\) into \(c(n)\)**:
Given:
\[
n(t) = -4t^2 + 100t
\]
\[
c(n) = 20000 + 5000n
\]
We can find \((c \circ n)(t)\) by substituting \(n(t)\) into \(c(n)\):
\[
(c \circ n)(t) = c(n(t)) = c(-4t^2 + 100t)
\]
Now we substitute \(n(t)\) into the cost function:
\[
(c \circ n)(t) = 20000 + 5000(-4t^2 + 100t)
\]
Simplifying this:
\[
(c \circ n)(t) = 20000 + 5000(-4t^2) + 5000(100t)
\]
\[
= 20000 - 20000t^2 + 500000t
\]
\[
= -20000t^2 + 500000t + 20000
\]

2. **Meaning of \((c \circ n)(t)\)**:
The function \((c \circ n)(t)\) represents the total cost of producing the number of automobile parts produced after \(t\) hours. Specifically, it gives the cost in dollars to produce the number of parts that can be produced in \(t\) hours, as expressed by the function \(n(t)\).

3. **Finding the domain**:
We begin by assessing the function \(n(t) = -4t^2 + 100t\). This is a parabola that opens downward.

- The number of parts produced \(n(t)\) must be non-negative because you cannot produce a negative quantity of parts. Therefore, we need to find when \(n(t) \geq 0\):
\[
-4t^2 + 100t \geq 0
\]
Factoring out \(-4\) gives:
\[
-4(t^2 - 25t) \geq 0 \quad \text{or} \quad t(t - 25) \leq 0
\]
The solutions to \(t(t - 25) = 0\) are \(t = 0\) and \(t = 25\). The inequality \(t(t - 25) \leq 0\) holds between these roots:
- Therefore, \(t \in [0, 25]\).

Thus, the domain of \((c \circ n)(t)\) is:
\[
\text{Domain: } [0, 25]
\]

In summary:
- The function \((c \circ n)(t)\) is given by \(c(n(t)) = -20000t^2 + 500000t + 20000\), representing the cost of producing car parts over time.
- The domain is the interval \([0, 25]\) hours, where production is valid.