Question 1173374: A small manufacturing makes and sells x machines each month. The monthly cost, C ,in dollars, of making x dollars is given by
C(x)= 0.35x^2 + 3200
The monthly revenue, R, in dollars, obtained by selling x machines is given by
R(x)= 180x - 0.55x^2
Find the smallest number of machines the company must make and sell each month in order to make a positive profit.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A small manufacturing makes and sells x machines each month.
The monthly cost, C ,in dollars, of making x dollars is given by
C(x)= 0.35x^2 + 3200
The monthly revenue, R, in dollars, obtained by selling x machines is given by
R(x)= 180x - 0.55x^2
Find the smallest number of machines the company must make and sell each month in order to make a positive profit.
:
Revenue > cost
180x - .55x^2 > .35x^2 + 3200
-.55x^2 - .35x^2 + 180x - 3200 > 0
-.90x^2 + 180x - 3200 > 0
the quadratic formula gives us two solutions
x > 180.3
and
x > 19.7
:
For least number of machines lets use x = 20 units
Revenue: 180(20) - .55*(20^2) = 3380
Cost: .35(20^2) + 3200 = 3060
a profit of $320
however is you use x=19 machines
Rev: 180(19) - .55(19^2) = 3221.45
Cost: .35(19^2) + 3200 = 3326.35
a loss of about $105
:
20 units the least to make a profit
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