Respuesta :
Here we have a problem of maximization and quadratic equations.
The unit prize that maximizes the revenue is $1,500, and the maximum revenue is $18,000.
We know that the revenue equation is:
R(P) = - 8p^2 + 24,000p
Where the variable p is the price.
Now we want to find the value of p that maximizes the revenue.
To do it, we can see that the revenue equation is a quadratic equation with a negative leading coefficient.
This means that the arms of the graph will go downwards, then the maximum point of the graph will be at the vertex.
Remember that for an equation like:
y = a*x^2 + b*x+ c
The x-value of the vertex is at:
x = -b/(2*a)
Then for the equation:
R(P) = - 8p^2 + 24,000p
The vertex is at:
p = -(24,000)/(2*-8) = 1,500
The value of p that maximizes the revenue is p = $1,500
To get the maximum revenue, we need to evaluate the revenue equation in that p value.
R(1,500) = - 8*(1,500)^2 + 24,000*1,500 = 18,000
And the revenue equation is in dollars, then the maximum revenue is 18,000 dollars.
If you want to learn more, you can read:
https://brainly.com/question/18269297
p = 1500 $ the unit price
R(p) = 18000000 $ maximum revenue
We will use two different procedures to calculate the maximum revenue.
That is equivalent to solve the problem and after that to test the solution
The first one is:
R(p) = - 8*p² + 24000*p
we realize that R(p) is a quadratic function ( a parabola) of the form:
y = a*x² + b*x + c ( c = 0 in this case)
We also know that as the coefficient of p² is negative the parabola opens downwards then the vertex is a maximum value for R(p), and the x coordinate of p is:
x = p = - b/2*a then by substitution
p = - ( 24000)/ 2 ( - 8)
p = 1500 $ and for that value of p
R(p) = - 8 ( 1500)² + 24000* (1500) = - 18000000 + 36000000
R(p) = 18000000 $
The second procedure is solving with the help of derivatives.
R(p) = - 8*p² + 24000*p
Tacking derivatives on both sides of the equation we get:
R´(p) = -16p + 24000
If R´(p) = 0 then -16p + 24000 = 0
p = 24000/ 16 p = 1500
if we check for the second derivative
R´´(p) = -16 -16 < 0 therefore there is a maximum value for R(p) when p = 1500, and that value is:
By substitution in R(p)
R(p) = -8 *(1500)² + 24000* 1500
R(p) = - 18000000 + 36000000
R(p) = 18000000 $
