Answer:
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
