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.
$\displaystyle R^{\prime}(x) = 80xe^{-0.4x} - 16x^2e^{-0.4x} = 0$ <-- Solve for x
$\displaystyle 80xe^{-0.4x} = 16x^2e^{-0.4x}$
$\displaystyle 5 = x$
(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...