Suppose the revenue (in thousands) for sales of x hundred units of an electronic item is given by the function R(x) = 40x^2e^-0.4x + 30, where the max capacity of the plant is eight hundred. Determine number of units for max revenue.
Suppose the revenue (in thousands) for sales of x hundred units of an electronic item is given by the function R(x) = 40x^2e^-0.4x + 30, where the max capacity of the plant is eight hundred. Determine number of units for max revenue.
<-- Solve for x
(I leave it to you to prove that x = 5 is a local maximum, not a local minimum.)
-Dan
Edit: This problem would actually belong to a Calculus forum...
Suppose the revenue (in thousands) for sales of x hundred units of an electronic item is given by the function R(x) = 40x^2e^-0.4x + 30, where the max capacity of the plant is eight hundred. Determine number of units for max revenue.
Originally Posted by madman1611
So you are saying the maximum number of units is 5, but the answer is 500 at back of book? Or am I totally confused. Thanks.
If you look at your original problem x is in units of 100.