2.) Maximizing revenue: The quantity demanded each month of the Sicard wristwatch is related to the unit price by the equation
p=50/0.01x^2 + 1 (0 < or = x < or = 20)
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?
revenue = price * demand
Assuming the usual carelessness with parentheses,
r(x) = x*p(x) = x(50/(0.01x^2 + 1))
The maximum revenue is found at x=10.
See the graph and other info at
http://www.wolframalpha.com/input/?i=x(50%2F(0.01x%5E2+%2B+1))
I do not understand how to get 10....
Do I just plot the graph filling in 1 2 3 etc for x?
To find the quantity of watches that must be sold to yield maximum revenue, we need to determine the value of x that maximizes the revenue function.
First, we need to express the revenue function in terms of x. Revenue is equal to the unit price multiplied by the quantity demanded, which is given by the equation p = 50/(0.01x^2 + 1). Therefore, the revenue function R(x) is:
R(x) = p * x
Substituting the expression for p in terms of x, we have:
R(x) = (50/(0.01x^2 + 1)) * x
R(x) = (50x)/(0.01x^2 + 1)
To find the maximum revenue, we need to find the value of x that maximizes R(x). We can do this by taking the derivative of R(x) with respect to x and setting it equal to zero. Let's find the derivative:
R'(x) = [(50(0.01x^2 + 1)) - (50x(0.02x))] / (0.01x^2 + 1)^2
R'(x) = [0.5x^2 + 50 - x^2] / (0.01x^2 + 1)^2
R'(x) = (49.5 - 0.5x^2) / (0.01x^2 + 1)^2
Setting R'(x) = 0, we have:
49.5 - 0.5x^2 = 0
0.5x^2 = 49.5
x^2 = 99
x = ±√99
Since x represents the quantity of watches sold, we can discard the negative value. Therefore, x = √99.
Finally, we need to check if this critical point gives a maximum. To do that, we can take the second derivative R''(x):
R''(x) = [1 - (0.01x^2)/(0.01x^2 + 1)^2] / (0.01x^2 + 1)^2
Evaluating R''(x) at x = √99, we find that R''(√99) > 0. This indicates that the critical point x = √99 corresponds to a maximum revenue.
Therefore, to yield maximum revenue, approximately √99 or about 9.95 thousand watches must be sold.