A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The materials for each feeder cost \( \$ 6 \), and the society sells an average of 28 per week at a price of \( \$ 10 \) each. The society has been considering raising the price, so it conducts a survey and finds that for every dollar increase, it loses 2 sales per week. (a) Find a function that models weekly profit in terms of price per feeder. (Let \( x \) represent the price per feeder and \( P \) represent the profit.) \( P(x)= \) What is the maximum weekly profit? (b) What price should the society charge for each feeder to maximize profits? \( \$ \) \$

Let's dive into creating a profit function for the bird-watching society! First up: the profits derived from the feeders! The profit \( P \) is calculated by taking the total revenue minus the total costs. The total revenue from selling \( q \) feeders at price \( x \) each is \( R(x) = x \cdot q \). Initially, they sell 28 feeders at \( x = 10\). For every dollar increase in price, sales drop by 2, so if they sell \( q \) feeders at price \( x \), we have: \[ q = 28 - 2(x - 10) = 68 - 2x \] Now, costs are pretty straightforward: total costs are \( 6q \), meaning: \[ C = 6(68 - 2x) = 408 - 12x \] Thus, the profit function becomes: \[ P(x) = R(x) - C = x(68 - 2x) - (408 - 12x) \] \[ = 68x - 2x^2 - 408 + 12x \] \[ = -2x^2 + 80x - 408 \] Now, let's find the maximum profit! The maximum profit of a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \): Here, \( a = -2 \) and \( b = 80 \): \[ x = -\frac{80}{2 \times -2} = 20 \] At \( x = 20 \): The number of feeders sold is \( q = 68 - 2 \times 20 = 28 \). Total revenue: \( R = 20 \times 28 = 560 \) Total cost: \( C = 6 \times 28 = 168 \) Thus, \( P = 560 - 168 = 392 \) So to wrap it up: (a) The profit function is \( P(x) = -2x^2 + 80x - 408 \). The maximum weekly profit is \( \$392 \). (b) The optimal price to maximize profits is \( \$20 \).