Suppose the function p= -1/8 x + 100

(0 =< x =<12) relates the selling
price p of an item to the quantity x
that is sold. Assume that p is in
dollars. What is the maximum
revenue possible in this situation?

To find the maximum revenue possible, we need to understand that revenue is calculated by multiplying the selling price by the quantity sold. In this case, the selling price is given by the function p = -1/8x + 100, and the quantity sold is represented by x.

Therefore, the revenue is calculated as R = p * x.

To find the maximum revenue, we can use calculus and find the critical points of the revenue function. The critical points occur where the derivative of the revenue function is equal to zero or undefined.

Step 1: Calculate the derivative of the revenue function:
R = p * x
dR/dx = dp/dx * x + p * dx/dx (using the product rule)
dR/dx = (-1/8) * x + 100 * 1 (since dx/dx = 1, and dp/dx = -1/8 according to the given function p = -1/8x + 100)

Step 2: Set the derivative equal to zero and solve for x:
(-1/8) * x + 100 = 0
(-1/8) * x = -100
x = (-100) / (-1/8)
x = 800

Step 3: Check if the critical point is within the given range (0 <= x <= 12):
Since 0 <= x <= 12, the critical point x = 800 is within the given range.

Step 4: Calculate the revenue at the critical point:
To find the revenue at the critical point, substitute x = 800 into the original revenue equation R = p * x:
R = (-1/8) * 800 + 100 * 800
R = -100 + 80000
R = 79900

Therefore, the maximum revenue possible in this situation is $79,900.

revenue is basically selling price - cost price , ignoring some other factors

I see an equation for the selling price, but no information about the cost.