Sure, let's go through each part of the question step-by-step for a clear understanding of how the given results are derived.
(a) Find the average rate of change (in \[tex]$ per unit) of \( C \) with respect to \( x \) when the production level is changed from \( x=100 \) to the given values.
The average rate of change of a function \( C(x) \) between two points \( x_1 \) and \( x_2 \) is calculated as:
\[ \text{Average Rate of Change} = \frac{C(x_2) - C(x_1)}{x_2 - x_1} \]
Given the cost function:
\[ C(x) = 1,000 + 9x + 0.15 x^2 \]
1. For \( x = 105 \):
- Initial production level, \( x_1 = 100 \)
- Final production level, \( x_2 = 105 \)
Calculate \( C(x) \) at \( x_1 \) and \( x_2 \):
\[ C(100) = 1,000 + 9(100) + 0.15(100)^2 \]
\[ C(100) = 1,000 + 900 + 1,500 = 2,400 \]
\[ C(105) = 1,000 + 9(105) + 0.15(105)^2 \]
\[ C(105) = 1,000 + 945 + 1,653.75 = 3,598.75 \]
Now, find the average rate of change:
\[ \frac{C(105) - C(100)}{105 - 100} = \frac{3,598.75 - 2,400}{5} = \frac{1,198.75}{5} = 239.75 \]
Therefore, the average rate of change when \( x \) changes from 100 to 105 is:
\[ \boxed{39.75} \text{ per unit} \]
2. For \( x = 101 \):
- Initial production level, \( x_1 = 100 \)
- Final production level, \( x_3 = 101 \)
Calculate \( C(x) \) at \( x_1 \) and \( x_3 \):
\[ C(100) = 2,400 \text{ (as previously calculated)} \]
\[ C(101) = 1,000 + 9(101) + 0.15(101)^2 \]
\[ C(101) = 1,000 + 909 + 1,530.15 = 3,439.15 \]
Now, find the average rate of change:
\[ \frac{C(101) - C(100)}{101 - 100} = \frac{3,439.15 - 2,400}{1} = 1039.15 \]
Therefore, the average rate of change when \( x \) changes from 100 to 101 is:
\[ \boxed{39.15} \text{ per unit} \]
(b) Find the instantaneous rate of change (in $[/tex] per unit) of [tex]\( C \)[/tex] with respect to [tex]\( x \)[/tex] when [tex]\( x = 100 \)[/tex].
The instantaneous rate of change of [tex]\( C(x) \)[/tex] at [tex]\( x = 100 \)[/tex] is found by taking the derivative of [tex]\( C(x) \)[/tex] and evaluating it at [tex]\( x = 100 \)[/tex].
Given the cost function:
[tex]\[ C(x) = 1,000 + 9x + 0.15 x^2 \][/tex]
The derivative [tex]\( C'(x) \)[/tex] is:
[tex]\[ C'(x) = \frac{d}{dx}(1,000 + 9x + 0.15 x^2) \][/tex]
[tex]\[ C'(x) = 0 + 9 + 0.30x \][/tex]
[tex]\[ C'(x) = 9 + 0.30x \][/tex]
Now, evaluate the derivative at [tex]\( x = 100 \)[/tex]:
[tex]\[ C'(100) = 9 + 0.30(100) \][/tex]
[tex]\[ C'(100) = 9 + 30 \][/tex]
[tex]\[ C'(100) = 39 \][/tex]
Therefore, the instantaneous rate of change at [tex]\( x = 100 \)[/tex] is:
[tex]\[ \boxed{39.00} \text{ per unit} \][/tex]
So, the answers are:
- For [tex]\( x = 105 \)[/tex]: [tex]\(\boxed{39.75} \)[/tex] per unit
- For [tex]\( x = 101 \)[/tex]: [tex]\(\boxed{39.15} \)[/tex] per unit
- Instantaneous rate of change at [tex]\( x = 100 \)[/tex]: [tex]\(\boxed{39.00} \)[/tex] per unit
