O.K. I found another Problem involving maximum revenue?

Question

Find the number of units, x, that produce a maximum revenue where

R(x) = 900x - 0.1x2

How would you go about solving this one and please show the results.

Answer 1

This being quadratic can be solved in 2 ways 1


algebra
1st make the cofficient of x^2 1 or a perfect quare
R = .1(9000x - x^2)
add and subtract 4500^2 to make it square(square of 1/2 of coffecient of x)
= .1(4500^2-4500^2+4500*2*x-x^2)
= .1(4500^2 -(4500-x)^2)
clearly maximum when 4500 =x then perfect square is minimum
putting this value you can find R(x)

second method
calculus
derivative of R(x) = 0(may be maximum or minimum)
900 -.2 x= 0 or x = 4500
because 2nd derivative is -ve so maxmimum

R(x) = 4500^2*.1 = 2025000

Answer 2

This is polynomial of 2nd degre p(x) = a*x^2 + b^x + c, with a= -0.1 and b = 900. Since a < 0 it has a maximum x* = -b/(2a) = -900/(-0.2) = 4500. Since this is an integer, it satisfies the given conditions,

Answer 3

R(x) = 900x - 0.1x2
R'(x)=900-.2x
set R'(x)=0
900=.2x
x=5*900=4500
R(4500)=900*4500-.1*4500^2=4050000 -2025000=2025000