q=D(x) = square root 340-x^2, where q = number of CDs sold per day when the price per cd is x dollars. At what price is the revenue a maximum
To find the price at which the revenue is a maximum, we need to find the derivative of the revenue function with respect to price, set it equal to zero, and solve for x.
First, let's define the revenue function R(x):
R(x) = x * D(x)
where D(x) is the number of CDs sold per day at a given price x.
We are given that D(x) = √(340 - x^2). Plugging this into the revenue function, we have:
R(x) = x * √(340 - x^2)
To find the maximum, we take the derivative of R(x) with respect to x:
R'(x) = √(340 - x^2) + x * 0.5 * (340 - x^2)^(-0.5) * (-2x)
Simplifying R'(x), we get:
R'(x) = √(340 - x^2) - x^2 / √(340 - x^2)
To find the critical points, we set R'(x) equal to zero:
0 = √(340 - x^2) - x^2 / √(340 - x^2)
To simplify, let's multiply both sides by √(340 - x^2) to get rid of the square root:
0 = (340 - x^2) - x^2
Combining like terms, we have:
0 = 340 - 2x^2
Solving for x^2, we get:
2x^2 = 340
x^2 = 170
Taking the square root of both sides, we have:
x = ±√170
Since x represents the price of a CD, we can ignore the negative solution. Therefore, the possible price at which revenue is a maximum is x = √170.
Thus, the revenue is a maximum when the price per CD is approximately √170 dollars.