Certainly! Let's go through this step-by-step.
### Part (a) Finding the Domain
The cost function is given as [tex]\( C = 6000 + 2x \)[/tex]. This function calculates the cost [tex]\( C \)[/tex] for producing [tex]\( x \)[/tex] units of a product.
1. Identify the given cost limits:
- The fixed initial cost is [tex]$6000.
- The variable cost per unit produced is $[/tex]2.
- The maximum cost is $10000.
2. Determine the domain of [tex]\( x \)[/tex]:
- The domain represents the possible values for [tex]\( x \)[/tex] (the number of units produced).
- Start with the minimum value for [tex]\( x \)[/tex]:
[tex]\[
x_{\text{min}} = 0
\][/tex]
This is because it is usually logical to assume that producing zero units is possible.
- Now, find the maximum value for [tex]\( x \)[/tex]:
- Set the cost function to the maximum cost:
[tex]\[
6000 + 2x = 10000
\][/tex]
- Solve for [tex]\( x \)[/tex]:
[tex]\[
2x = 10000 - 6000
\][/tex]
[tex]\[
2x = 4000
\][/tex]
[tex]\[
x = \frac{4000}{2} = 2000
\][/tex]
- So, the maximum value for [tex]\( x \)[/tex] is 2000.
3. Express the domain:
- Combining these results, the domain is:
[tex]\[
0 \leq x \leq 2000
\][/tex]
### Part (b) Finding the Range
The range of a function represents the possible values that the output (in this case, [tex]\( C \)[/tex]) can take.
1. Determine the minimum value of [tex]\( C \)[/tex]:
- The minimum cost occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[
C_{\text{min}} = 6000 + 2 \cdot 0 = 6000
\][/tex]
2. Determine the maximum value of [tex]\( C \)[/tex]:
- The maximum cost corresponds to the maximum [tex]\( x \)[/tex]:
[tex]\[
x = 2000
\][/tex]
- Using the cost function:
[tex]\[
C_{\text{max}} = 6000 + 2 \cdot 2000 = 6000 + 4000 = 10000
\][/tex]
3. Express the range:
- Combining these results, the range is:
[tex]\[
6000 \leq C \leq 10000
\][/tex]
### Final Answer
Putting all this information together:
(a) The domain is:
[tex]\[
0 \leq x \leq 2000
\][/tex]
(b) The range is:
[tex]\[
6000 \leq C \leq 10000
\][/tex]
This concludes the detailed step-by-step solution to the problem.
