Find the number of units that produces a
maximum revenue given by R = 290x−0.2x
2, where R is the total revenue in dollars and x is the number of units sold.
Answer in units of units.
A quadratic equation have the standrad form:
a x ^ 2 + b x + c
R = 290 x − 0.2 x ^ 2 is quadratic equation:
R = − 0.2 x ^ 2 + 290 x + 0
The vertex of a quadratic parabola is the highest or lowest point, the maximum or minimum.
When a < 0 a quadratic parabola has a maximum.
In this equation : a = - 0.2 , b = 290 , c = 0
x coordinate of vertex:
x = - b / 2 a
In this case:
x = - 290 / [ 2 * ( - 0.2 ) ]
x = - 290 / - 0.4
x = 725
Replace this value in equation:
R = 290 x − 0.2 x ^ 2
Rmax = 290 * 725 - 0.2 * 725 ^ 2
Rmax = 210,250 - 0.2 * 525,625
Rmax = 210,250 - 105,125
Rmax = 105,125
725 units sold
Rmax = $105,125
To find the number of units that produces a maximum revenue, we need to maximize the revenue function R = 290x - 0.2x^2.
To do this, we can take the derivative of the revenue function with respect to x and set it equal to zero, and solve for x.
Differentiating R with respect to x:
dR/dx = 290 - 0.4x
Setting the derivative equal to zero:
290 - 0.4x = 0
Solving for x:
0.4x = 290
x = 290 / 0.4
x = 725
Therefore, the number of units that produces a maximum revenue is 725 units.
To find the number of units that produces the maximum revenue, we need to find the value of x that maximizes the revenue function R.
Given the revenue function: R = 290x - 0.2x^2
To find the maximum, we can either take the derivative of the revenue function and set it equal to zero, or we can graph the function and identify the maximum point.
Let's first take the derivative and set it equal to zero:
dR/dx = 290 - 0.4x
Setting the derivative equal to zero:
290 - 0.4x = 0
Solving for x:
0.4x = 290
x = 290 / 0.4
x = 725
So, the number of units that produces the maximum revenue is 725 units.