Certainly! Let's address both parts of the question step-by-step.
### Part (A): Exact Profit from the Sale of the 26th Machine
The given profit function is:
[tex]\[ P(x) = 20x - 0.5x^2 - 270 \][/tex]
To find the exact profit from the sale of the 26th machine, we evaluate this function at [tex]\( x = 26 \)[/tex].
[tex]\[ P(26) = 20(26) - 0.5(26)^2 - 270 \][/tex]
Breaking it down:
1. Calculate [tex]\( 20 \times 26 \)[/tex]:
[tex]\[ 20 \times 26 = 520 \][/tex]
2. Calculate [tex]\( 0.5 \times 26^2 \)[/tex]:
[tex]\[ 26^2 = 676 \][/tex]
[tex]\[ 0.5 \times 676 = 338 \][/tex]
3. Substitute these into the function:
[tex]\[ P(26) = 520 - 338 - 270 \][/tex]
4. Simplify the expression:
[tex]\[ P(26) = 520 - 338 = 182 \][/tex]
[tex]\[ P(26) = 182 - 270 = -88 \][/tex]
So, the exact profit from the sale of the 26th machine is:
[tex]\[ \boxed{-88} \][/tex]
### Part (B): Approximate Profit Using Marginal Profit
The marginal profit is the derivative of the total profit function [tex]\( P(x) \)[/tex]. First, we need to compute the derivative of [tex]\( P(x) \)[/tex]:
Given:
[tex]\[ P(x) = 20x - 0.5x^2 - 270 \][/tex]
The derivative [tex]\( P'(x) \)[/tex] (marginal profit function) is:
[tex]\[ P'(x) = \frac{d}{dx}(20x - 0.5x^2 - 270) \][/tex]
To get the marginal profit:
1. The derivative of [tex]\( 20x \)[/tex] is [tex]\( 20 \)[/tex]
2. The derivative of [tex]\( -0.5x^2 \)[/tex] is [tex]\( -x \)[/tex]
3. The derivative of the constant [tex]\( -270 \)[/tex] is [tex]\( 0 \)[/tex]
So,
[tex]\[ P'(x) = 20 - x \][/tex]
Now, we evaluate [tex]\( P'(x) \)[/tex] at [tex]\( x = 26 \)[/tex]:
[tex]\[ P'(26) = 20 - 26 = -6 \][/tex]
Therefore, the approximate profit from the sale of the 26th machine using the marginal profit is:
[tex]\[ \boxed{-6} \][/tex]
In summary:
- Exact profit on the 26th machine: [tex]\( \boxed{-88} \)[/tex]
- Approximate profit on the 26th machine: [tex]\( \boxed{-6} \)[/tex]
