A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The
materials for each feeder cost , and the society sells an average of 28 per week at a price of 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 represent the price per feeder and
represent the profit.)
What is the maximum weekly profit?
(b) What price should the society charge for each feeder to maximize profits?
$

Ask by Klein Macdonald. in the United States

Mar 19,2025

Question

A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The
materials for each feeder cost , and the society sells an average of 28 per week at a price of 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 represent the price per feeder and
represent the profit.)
What is the maximum weekly profit?
(b) What price should the society charge for each feeder to maximize profits?
$

Ask by Klein Macdonald. in the United States

Mar 19,2025

UpStudy AI · Accepted Answer

Let $$ x $$ be the price per feeder. The society’s original situation is a price of $$ \$10 $$ with 28 feeders sold per week. For every additional dollar, they lose 2 sales. Thus, for a price $$ x $$:

- The number of feeders sold per week is  
  $$
  28 - 2(x-10).
  $$

Each feeder has a material cost of $$ \$6 $$; hence the profit per feeder is  
$$
x - 6.
$$

The weekly profit function $$ P(x) $$ is the product of the profit per feeder and the number of feeders sold:
$$
P(x) = (x - 6)\Bigl(28 - 2(x-10)\Bigr).
$$

Simplify the second factor:
$$
28 - 2(x-10) = 28 - 2x + 20 = 48 - 2x.
$$

Thus, the profit function becomes:
$$
P(x)= (x-6)(48-2x).
$$

Next, expand the product:
$$
P(x)= (x-6)(48-2x)= 48x - 2x^2 - 288 + 12x = -2x^2 + 60x - 288.
$$

This quadratic function, $$ P(x) = -2x^2 + 60x - 288 $$, opens downward (since the coefficient of $$ x^2 $$ is negative), meaning its vertex gives the maximum profit.

The $$ x $$-coordinate of the vertex is found by:
$$
x = -\frac{b}{2a} = -\frac{60}{2(-2)} = 15.
$$

Thus, the maximum weekly profit occurs when the price per feeder is $$ \$15 $$.

Now, calculate the maximum profit:
$$
P(15)= -2(15)^2 + 60(15) - 288.
$$

Step by step:
$$
-2(15)^2 = -2(225) = -450,
$$
$$
60(15)= 900,
$$
$$
P(15)= -450 + 900 - 288 = 450 - 288 = 162.
$$

Therefore, the maximum weekly profit is $$ \$162 $$.

$$
\boxed{P(x)= (x-6)(48-2x), \quad \text{maximum profit is } \$162, \quad \text{attained at } x=\$15.}
$$
The weekly profit function is $$ P(x) = (x - 6)(48 - 2x) $$. The maximum weekly profit is \$162, achieved when the price per feeder is \$15.

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