Asked by pypski
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?
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?
Answers
Answered by
Steve
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))
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))
Answered by
pypski
I do not understand how to get 10....
Answered by
pypski
Do I just plot the graph filling in 1 2 3 etc for x?
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