To solve this problem, consider the given total cost function [tex]\( C(x) = 0.01 x^2 + 0.4 x + 30 \)[/tex], where [tex]\( x \)[/tex] is the number of units produced and sold.
We need to find two things:
1. [tex]\(\Delta C\)[/tex] when [tex]\( x = 90 \)[/tex] and [tex]\( \Delta x = 1 \)[/tex]
2. The derivative [tex]\( C'(x) \)[/tex] evaluated at [tex]\( x = 90 \)[/tex]
### Step 1: Calculate [tex]\(\Delta C\)[/tex]
Given [tex]\( x = 90 \)[/tex] and [tex]\( \Delta x = 1 \)[/tex]:
[tex]\[
\Delta C = C(x + \Delta x) - C(x)
\][/tex]
Substitute [tex]\( x \)[/tex]:
[tex]\[
\Delta C = C(90 + 1) - C(90)
\][/tex]
Calculate [tex]\( C(90) \)[/tex]:
[tex]\[
C(90) = 0.01 \times (90)^2 + 0.4 \times 90 + 30
\][/tex]
[tex]\[
C(90) = 0.01 \times 8100 + 36 + 30
\][/tex]
[tex]\[
C(90) = 81 + 36 + 30
\][/tex]
[tex]\[
C(90) = 147
\][/tex]
Calculate [tex]\( C(91) \)[/tex]:
[tex]\[
C(91) = 0.01 \times (91)^2 + 0.4 \times 91 + 30
\][/tex]
[tex]\[
C(91) = 0.01 \times 8281 + 36.4 + 30
\][/tex]
[tex]\[
C(91) = 82.81 + 36.4 + 30
\][/tex]
[tex]\[
C(91) = 149.21
\][/tex]
Then:
[tex]\[
\Delta C = C(91) - C(90) = 149.21 - 147
\][/tex]
[tex]\[
\Delta C = 2.21
\][/tex]
### Step 2: Calculate the derivative [tex]\( C'(x) \)[/tex]
The derivative of the cost function:
[tex]\[
C(x) = 0.01 x^2 + 0.4 x + 30
\][/tex]
[tex]\[
C'(x) = 0.02 x + 0.4
\][/tex]
Evaluate [tex]\( C'(x) \)[/tex] at [tex]\( x = 90 \)[/tex]:
[tex]\[
C'(90) = 0.02 \times 90 + 0.4
\][/tex]
[tex]\[
C'(90) = 1.8 + 0.4
\][/tex]
[tex]\[
C'(90) = 2.2
\][/tex]
### Final Results
[tex]\[
\Delta C = \$2.21
\][/tex]
[tex]\[
C'(x) \text{ when } x = 90 \text{ is } \$2.2 \text{ per unit}
\][/tex]
