SOLUTION: A cell phone company has the following cost and revenue functions: C(x) = 8x^2 − 600x + 21,500 and R(x) = −3x^2 + 480x What is the quantity range of cell phones they should

Algebra ->  Systems-of-equations -> SOLUTION: A cell phone company has the following cost and revenue functions: C(x) = 8x^2 − 600x + 21,500 and R(x) = −3x^2 + 480x What is the quantity range of cell phones they should       Log On


   



Question 1138659: A cell phone company has the following cost and revenue functions:
C(x) = 8x^2 − 600x + 21,500 and
R(x) = −3x^2 + 480x
What is the quantity range of cell phones they should produce each day so there is a profit? Round to the nearest numbers that generate profit.

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Look at R(x)-C(x) for profit
That is -3x^2+480x-(8x^2-600x+21500)=
watch signs
-11x^2+1080x-21500>0, to generate a profit.
The maximum is at x=-b/2a=-1080/-22=49.09
the zero point is found using the quadratic formula. The discriminant b^2-4ac is 220400, and its square root is 469.47
x=(-1080+/-469.47)/-22, so one root: (-1080-469.47/-22)=70.43 or 70 and the other is -1080+469.47/-22=27.75 or 28
The numbers are between 28 and 70 inclusive. ANSWER
graph%28300%2C300%2C-10%2C80%2C-20000%2C5000%2C-11x%5E2%2B1080x-21020%29