revenues???

Question

The weekly revenue R, in dollars, from selling x calculators is

R(x) = -20x^2 + 1000x

a. Determine where the graph of R is increasing and where it is decreasing.


b. How many calculators need to be sold to maximize revenue?


c. What is the maximum revenue?

Answer 1

R(x) = -20x^2 + 1000x
R'(x) = -40x + 1000; critical point at x=25. If x<25 R' >0, and if x>25 R'<0.

R'(x)=0 at x=25, which provides max revenue.

Max rev = -20*(25^2)+1000*25=12500

Answer 2

Because this is a quadratic function, you can determine the increasing/decreasing intervals and maximum revenue by calculating the vertex. Use x = -b/2a to find the # of calculators to be sold, then plug that number in to find the revenue.
x = -1000/(2*-20) = 25
so

a) increasing for x < 25 and decreasing for x > 25.

b) 25 calculators will bring in max. revenue

c) R(25) = -20 * 25^2 + 1000 * 25 = $12,500

Answer 3

R(x) = -20x^2 + 1000x
R(x) = -20(x^2 - 50x)
R(x) = -20(x^2 - 50x + 25^2 - 25^2)
R(x) = -20(x - 25)^2 + 12,500
a.
R increases for x < 25
R decreases for x > 25
b. R is maximum at x = 25
c. $12,500