Show all steps and Find the maximum revenue for the revenue function R(x)= 360x - 0.05x^2

R' = 360-.1x

R'=0 when x = 3600
R(3600) = 648,000

How did you .1x? and where did the 3600 come from?

ok. sorry. I forgot this was algebra, not calculus.

recall that the vertex of ax^2+bx+c=0 is at x = -b/2a

R(x) is a parabola, with vertex at x = 360/0.1
R(3600) = 648,000

To find the maximum revenue for the revenue function R(x) = 360x - 0.05x^2, we can use the concept of finding the vertex of a quadratic function. The vertex of a quadratic function is the point where the function reaches its maximum or minimum value.

Step 1: Rewrite the function in vertex form
To find the vertex, we need to rewrite the equation in vertex form, which is of the form: y = a(x-h)^2 + k, where (h, k) represents the coordinates of the vertex.

Let's rewrite the revenue function in vertex form:
R(x) = -0.05x^2 + 360x

Step 2: Identify the coefficients in the equation
In our equation, a = -0.05 and h = -b/2a, where b is the coefficient of x. In this case, b = 360.

Step 3: Calculate h
Using the formula h = -b/2a, we can calculate the value of h:
h = -360 / (2 * -0.05)
h = -360 / -0.1
h = 3600

Step 4: Substitute h into the equation to find k
Substituting the value of h into the original equation, we can solve for k:
R(x) = -0.05x^2 + 360x
R(3600) = -0.05(3600)^2 + 360(3600)
R(3600) = -0.05(12,960,000) + 1,296,000
R(3600) = -648,000 + 1,296,000
R(3600) = 648,000

Step 5: Determine the maximum revenue
The vertex of the function is (h, k), so the maximum revenue occurs when x = 3600, and the maximum revenue is 648,000.

Therefore, the maximum revenue for the revenue function R(x) = 360x - 0.05x^2 is 648,000.