The maximum profit is $288, and you should charge $6 per dozen.
Let's assume that demand depends linearly on price, such that we can write:
q = a*p + b
where q is the demand and p is the price.
We have 3 points of that line:
The slope of the line is given by:
[tex]a = \frac{q_2 - q_1}{p_2 - p_1}[/tex]
Where these are values of demand and prize for two different points, if we use the first two points we get:
[tex]a = \frac{72 - 80}{3 - 2} = -8[/tex]
Then the line is something like:
q = -8*p + b
Remember that when p = 2, the demand is equal to 80, replacing that we get:
80 = -8*2 + b
80 + 16 = b = 96.
So the line is:
q = -8*p + b
Profit is given by the product between the demand and the price, so we have.
Profit = p*q = p*(-8*p + 96) = -8p^2 + 96p
We want to maximize that, notice that is a quadratic equation of negative leading coefficient, so the maximum is at the vertex, using the equation for the vertex of a quadratic equation we will see that the maximum is at:
p = -96/(2*-8) = 6
Then the maximum profit is:
profit = -8*(6)^2 + 96*6 = 288
The maximum profit is $288, and you should charge $6 per dozen.
If you want to learn more about linear relations, you can read:
https://brainly.com/question/1884491
