Answer:
The required function is: 
The price for each feeder to maximize profit should be $13.
The maximum weekly profit is $147
Step-by-step explanation:
Consider the provided information.
Part (A)
The materials for each feeder cost $6, and the society sells an average of 30 per week at a price of $10 each.
Let x be the price per feeder.
Profit = Revenue - cost
So the profit per feeder is 
If we increase then for every dollar increase, it loses 3 sales per week.
This can be written as:



Where x-10 represents the increase in price and 3 represents the decrease in sales per week.
Thus the profit will be:


Hence, the required function is: 
Part (B) What price should the society charge for each feeder to maximize profits? What is the maximum weekly profit?
The above function is a downward parabola so the maximum will occur at vertex.
The x coordinate of the vertex of parabola is: 
Substitute a=-3, b=78 and c=-360 in above.

Hence, the price for each feeder to maximize profit should be $13.
Now substitute the value of x in 

Hence, the maximum weekly profit is $147