g A gift shop sells 40 wind chimes per month at $110 each. The owners estimate that for each $11 increase in price, they will sell 2 fewer wind chimes per month. Find the price per wind chime that will maximize revenue.

The price that maximizes the revenue is $165

Step-by-step explanation:

We can model the price as a function of sold units as a linear relationship.

Remember that a linear relationship is something like:

y = a*x + b

where a is the slope and b is the y-intercept.

We know that if the line passes through the points (x₁, y₁) and (x₂, y₂), then the slope can be written as:

a = (y₂ - y₁)/(x₂ - x₁)

For this line, we have the point (40, $110)

which means that to sell 40 units, the price must be $110

And we know that if the price increases by $11, then he will sell 2 units less.

Then we also have the point (38, $121)

So we know that our line passes through the points (40, $110) and  (38, $121)

Then the slope of the line is:

a = ($121 - $110)/(38 - 40) = $11/-2 = -$5.5

Then the equation of the line is:

p(x) = -$5.5*x + b

to find the value of b, we can use the point (40, $110)

This means that when x = 40, the price is $110

then:

p(40) = $110 = -$5.5*40 + b

            $110 = -$220 + b

       $110 + $220 = b

        $330 = b

Then the price equation is:

p(x) = -$5.5*x + $330

Now we want to find the maximum revenue.

The revenue for selling x items, each at the price p(x), is:

revenue = x*p(x)

replacing the p(x) by the equation we get:

revenue = x*(-$5.5*x + $330)

revenue = -$5.5*x^2 + $330*x

Now we want to find the x-value for the maximum revenue.

You can see that the revenue equation is a quadratic equation with a negative leading coefficient. This means that the maximum is at the vertex.

And remember that for a quadratic equation like:

y = a*x^2 + b*x + c

the x-value of the vertex is:

x = -b/2a

Then for our equation:

revenue = -$5.5*x^2 + $330*x

the x-vale of the vertex will be:

x = -$330/(2*-$5.5) = 30

x = 30

This means that the revenue is maximized when we sell 30 units.

And the price is p(x) evaluated in x = 30

p(30) = -$5.5*30 + $330 = $165

The price that maximizes the revenue is $165

maybe i can

126 maybe. sorry