R' = 360-.1x
R'=0 when x = 3600
R(3600) = 648,000
Show all steps and Find the maximum revenue for the revenue function R(x)= 360x - 0.05x^2
3 answers
How did you .1x? and where did the 3600 come from?
ok. sorry. I forgot this was algebra, not calculus.
recall that the vertex of ax^2+bx+c=0 is at x = -b/2a
R(x) is a parabola, with vertex at x = 360/0.1
R(3600) = 648,000
recall that the vertex of ax^2+bx+c=0 is at x = -b/2a
R(x) is a parabola, with vertex at x = 360/0.1
R(3600) = 648,000
1 answer