Using the equations R(x)=-x^2+400x and C(x)=x^2+40x+100, find the Maximum Revenue
P(x) = R(x) - C(x)
P(x) =-x^2 +400x -x^2 -40x -100
P(x) -2x^2 + 360x -100
P'(x) = -2x +360
solve for x
0 = -2x + 360
P(x) = R(x) - C(x)
P(x) =-x^2 +400x -x^2 -40x -100
P(x) -2x^2 + 360x -100
P'(x) = -4x +360
solve for x
0 = -4x + 360
R'(x ) -2x + 400
0 = -2x + 400
2x = 400
x = 200
R(200) = -(200)^2 + 400(200)
-40000 + 80000 = 40000
To find the maximum revenue, we need to determine the value of x that maximizes the revenue function.
The revenue function, R(x), is given by R(x) = -x^2 + 400x, where x represents the quantity of the item being sold.
To find the maximum revenue, we can use calculus. We need to take the derivative of R(x) and set it equal to zero to find the critical points. Then, we can determine the second derivative at those critical points to check for the maximum.
1. Find the derivative of R(x):
R'(x) = -2x + 400
2. Set the derivative equal to zero:
-2x + 400 = 0
-2x = -400
x = 200
So, x = 200 is a critical point.
3. Determine the second derivative of R(x):
R''(x) = -2
4. Substitute the critical point (x = 200) into the second derivative:
R''(200) = -2
Since the second derivative is negative, this indicates that x = 200 is a local maximum.
Now, let's calculate the maximum revenue.
Plug the critical point (x = 200) into the revenue function R(x):
R(200) = -(200)^2 + 400(200)
R(200) = -40000 + 80000
R(200) = 40000
Therefore, the maximum revenue is $40,000.