Find the maximum revenue using the following equations: R(x)=-x^2 +400x and C(x)=x^2+40x+100.

What I've done so far is use R(x) and solve for x, which got me 0 and 400. Is that how you're supposed to start it? What happens next? (I apologize; I'm really bad at teaching myself how to work problems without a teacher.)

max R when R'=0

R' = -2x+400
R'=0 when x=200

In a way, you are on the right track. R is a parabola, whose vertex is midway between the roots. That is, at x=200. Using R' always works, though, even when R is not a nice easy function like a quadratic. This is a calculus class. All these max/min problems involve the derivative, so get used to that idea.

To find the maximum revenue, you need to find the value of x that maximizes the revenue function R(x). It seems like you have already found the critical points of R(x) by setting its derivative equal to zero and solving for x. Well done!

Now, to determine which critical point yields the maximum revenue, you can use the second derivative test. The second derivative of R(x) will help us determine if the critical points are maximum points, minimum points, or neither.

Let's start by finding the second derivative of R(x). The derivative of R(x) with respect to x gives us the rate of change of revenue, which is given by the slope of the revenue function. So, let's calculate the derivative:

R'(x) = -2x + 400

Now, taking the derivative of R'(x) will give us the second derivative of R(x):

R''(x) = -2

Since the second derivative R''(x) is always negative (-2 in this case), it indicates that the revenue function R(x) is concave downward, and the critical points are maximum points. Thus, x = 0 and x = 400 are both points of maximum revenue.

Now, you need to substitute these critical points back into the revenue function R(x) to find the maximum revenue.

For x = 0:
R(0) = -(0)^2 + 400(0) = 0

For x = 400:
R(400) = -(400)^2 + 400(400) = -160,000 + 160,000 = 0

As you can see, both x = 0 and x = 400 yield a revenue of 0. However, note that this means your revenue is maximized at 0, not 400. This indicates a somewhat unusual situation, where you have no revenue at the maximum point. Most likely, there's an error in the given revenue function R(x), or additional context is needed to make sense of the situation.

Regarding the cost function C(x), it is not directly involved in finding the maximum revenue. However, it can be useful to find the corresponding values of the cost at the maximum revenue points, which can help evaluate the profitability of the business. You can substitute x = 0 and x = 400 into the cost function C(x) to find the corresponding costs.