Suppose the function p= -1/8 x + 100
(0 grtr&= 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 in this situation, we need to understand that revenue is calculated by multiplying the selling price (p) by the quantity sold (x).
The given function p = -1/8x + 100 represents the selling price (p) in terms of the quantity sold (x). Notice that the coefficient in front of x is negative, indicating a decreasing linear relationship between p and x.
To maximize revenue, we need to find the value of x that gives the highest possible value for p, as this will result in the highest revenue.
To do this, we can differentiate the function p = -1/8x + 100 with respect to x and set it to zero to find the critical point:
dp/dx = 0
Differentiating the function, we get:
dp/dx = -1/8
Setting dp/dx = 0, we find:
-1/8 = 0
However, this equation has no solution as there is no value of x that makes the derivative equal to zero. This implies that there is no maximum revenue in this situation.
Instead, we can analyze the function to understand the behavior of p with changing values of x. Since the coefficient of x is negative, as x increases, p decreases. This means that the maximum revenue occurs when the quantity sold (x) is at its minimum, which is 0 in this case.
Therefore, the maximum revenue possible in this situation is when no items are sold (x = 0). Substituting x = 0 into the function, we find:
p = -1/8(0) + 100
p = 100
Hence, the maximum revenue in this situation is $100, which occurs when no items are sold.