SOLUTION: The total annual revenue for a product is given by R(x) = 10,000x-2x^2 dollars, where x is the number of units sold. (a) To maximize revenue, how many units must be sold? (b) Fi

Algebra ->  Finance -> SOLUTION: The total annual revenue for a product is given by R(x) = 10,000x-2x^2 dollars, where x is the number of units sold. (a) To maximize revenue, how many units must be sold? (b) Fi      Log On


   

Question 1195777: The total annual revenue for a product is given by R(x) = 10,000x-2x^2
dollars, where x is the number of units sold.
(a) To maximize revenue, how many units must be sold?
(b) Find the maximum possible annual revenue?
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!


R(x) = 10000x-2x%5E2+

dR/dx = 10000-4x



At min or max, dR/dx = 0:  10000-4x = 0  ==>  x = 2500 units for max revenue. 

You can see this is a max (and not a min) by looking at 2nd deriv. which is -4 ==> concave down ==> critical point is a local maximum.



Plug in x=2500 into R(x) to get the max revenue.  To convince yourself, you should plug in x=2501 and x=2499 to see those R(x) values are less than R(2500).