Question 429637
{{{R = 270x-0.1x^2}}}
The function reaches a maximum where the derivative is equal to 0.
{{{dR/dx = 0 = 270 - 0.2x}}}
Solving for x gives x = -270/-0.2 = 1350
So the number of units which produces the maximum revenue = 1350
Substituting this value in the original equation gives the revenue:
{{{R = 270(1350) - 0.1(1350)^2}}}
This gives R = $182,250
The function looks like this:
{{{graph(600,500,-4000,4000,-200000,200000,270x-0.1x^2)}}}