Find the number of units that produces a maximum revenue. The revenue R is measured in dollars and x is the number of units produced.
R = 80x – 0.0001x^2
as with all parabolas, the vertex (in this case a maximum) is at
x = -b/2a = 80/0.0002
Ah, the quest for maximum revenue! Well, my dear friend, to find the number of units that will deliver this magical maximum revenue, we need to follow a particular path.
First, let's examine the equation R = 80x – 0.0001x^2. This equation gives us the revenue R in dollars as a function of the number of units produced, x.
To find the maximum revenue, we need to determine the value of x that will make R as large as possible. How do we do that, you might ask? Well, we can use a little bit of mathematical magic called differentiation!
By taking the derivative of the revenue function with respect to x (dR/dx), we can find the critical points where the maximum revenue could occur. So, let's put on our math hats and work it out:
dR/dx = 80 - 0.0002x
Now, we set this derivative equal to zero to find the critical point:
80 - 0.0002x = 0
Solving for x, we have:
x = 80/0.0002
x = 400,000
Voilà! We have found a possible critical point. But hold your horses, my friend! We need to confirm that this point indeed corresponds to a maximum revenue.
To do that, we can take the second derivative of the revenue function, which is d^2R/dx^2:
d^2R/dx^2 = -0.0002
Since the second derivative is negative, it tells us that we are dealing with a concave down function, which means that the critical point we found corresponds to a maximum.
Therefore, the number of units that produces a maximum revenue is x = 400,000.
But hey, don't take my word for it, go forth and double-check my calculations! After all, even the best of us clowns can make a mathematical misstep every now and then.
To find the number of units that produces a maximum revenue, we need to maximize the revenue function R = 80x - 0.0001x^2 with respect to x.
To do this, we can take the derivative of R with respect to x and set it equal to zero, since the critical points (where the derivative is zero) can potentially correspond to maximum or minimum points.
Let's find the derivative of R:
dR/dx = 80 - 0.0002x
Setting this equal to zero and solving for x:
80 - 0.0002x = 0
0.0002x = 80
x = 80 / 0.0002
x = 400,000
So, the critical point is x = 400,000 units.
Now, we need to determine whether this point corresponds to a maximum or minimum value for R. We can use the second derivative test to check this.
Let's find the second derivative of R:
d^2R/dx^2 = -0.0002
Since the second derivative is negative, this means the point x = 400,000 corresponds to a maximum value for R.
Therefore, the number of units that produces a maximum revenue is 400,000 units.