Let's address the given problem step by step:
### Part A: Finding the Marginal Profit Function
The profit function provided is:
[tex]\[ P(q) = -0.02q^2 + 4q - 26 \][/tex]
To find the marginal profit function, we need to take the first derivative of the profit function [tex]\( P(q) \)[/tex] with respect to [tex]\( q \)[/tex].
[tex]\[ \frac{dP(q)}{dq} = \frac{d}{dq}(-0.02q^2 + 4q - 26) \][/tex]
Using basic differentiation rules:
[tex]\[ \frac{d}{dq}(-0.02q^2) = -0.04q \][/tex]
[tex]\[ \frac{d}{dq}(4q) = 4 \][/tex]
[tex]\[ \frac{d}{dq}(-26) = 0 \][/tex]
Combining these results:
[tex]\[ MP(q) = -0.04q + 4 \][/tex]
So, the marginal profit function is:
[tex]\[ MP(q) = 4 - 0.04q \][/tex]
### Part B: Finding the Quantity of Sunglasses to Maximize Profits
To maximize profits, we need to find the critical points of the profit function by setting the marginal profit function equal to zero:
[tex]\[ 4 - 0.04q = 0 \][/tex]
Solving for [tex]\( q \)[/tex]:
[tex]\[ 0.04q = 4 \][/tex]
[tex]\[ q = \frac{4}{0.04} \][/tex]
[tex]\[ q = 100 \][/tex]
Therefore, 100 thousand pairs of sunglasses need to be sold to maximize profits.
### Part C: Calculating the Maximum Profit
To find the maximum profit, we evaluate the original profit function [tex]\( P(q) \)[/tex] at [tex]\( q = 100 \)[/tex]:
[tex]\[ P(100) = -0.02(100)^2 + 4(100) - 26 \][/tex]
Calculating each term:
[tex]\[ -0.02(100)^2 = -0.02(10000) = -200 \][/tex]
[tex]\[ 4(100) = 400 \][/tex]
[tex]\[ -26 = -26 \][/tex]
Combining these results:
[tex]\[ P(100) = -200 + 400 - 26 = 174 \][/tex]
So, the maximum profit is 174 thousand dollars.
### Summary of Results
A) The simplified expression for the marginal profit function is:
[tex]\[ MP(q) = 4 - 0.04q \][/tex]
B) To maximize profits, 100 thousand pairs of sunglasses need to be sold.
C) The maximum expected profit is 174 thousand dollars.
Feel free to ask for further clarifications if needed!
