The quantity demanded each month of the Sicard wristwatch is related to the unit price given by the equation below, where p is measured in dollars and x is measured in units of a thousand. To yield a maximum revenue, how many watches must be sold? (Round your answer to the nearest whole number.)
p = 50/(0.01 x^2 + 1) (0<=x<=20)
watches
To find the number of watches that must be sold to yield a maximum revenue, we need to understand the relationship between the quantity demanded and the unit price. The equation given is:
p = 50/(0.01x^2 + 1)
where p represents the unit price in dollars and x represents the quantity demanded in units of a thousand.
To maximize revenue, we need to find the point at which the price times the quantity demanded is maximum. This can be calculated by multiplying the price equation by x:
px = 50x/(0.01x^2 + 1)
Now, we can find the derivative of the revenue function with respect to x:
d(px)/dx = 50(0.01x^2 + 1) - 50x(0.02x) / (0.01x^2 + 1)^2
Setting the derivative equal to zero to find the critical point:
50(0.01x^2 + 1) - 50x(0.02x) / (0.01x^2 + 1)^2 = 0
Simplifying the equation:
0.5x^2 + 50 - x^2 / (0.01x^2 + 1)^2 = 0
0.5x^2 + 50 = x^2 / (0.01x^2 + 1)^2
(0.5x^2 + 50)(0.01x^2 + 1)^2 = x^2
Expanding and rearranging the equation:
0.01x^4 + 0.02x^2 + x^2 + 0.5x^2 + 50 - x^2 = 0
0.01x^4 + 1.5x^2 + 50 = 0
Solving this quadratic equation is quite complex and may require numerical methods or advanced mathematical techniques. However, for the given range of x (0<=x<=20), we can find the maximum by evaluating the revenue function at x=0 and x=20, and then finding the maximum among these values.
Substituting x=0:
p(0) = 50/(0.01(0)^2 + 1) = 50
Substituting x=20:
p(20) = 50/(0.01(20)^2 + 1) = 50/(0.01(400) + 1) = 50/(4 + 1) = 50/5 = 10
Thus, the maximum revenue is obtained when x=20, which means 20,000 watches must be sold.