equation word problem

Question

A company manufactures small flash drives for home computers. It knows from experience that the number of flash drives it can sell each day, x, is related to the price p per flash drive by the equation x = 800 - 100p. At what price should the company sell the flash drives if it wants the daily revenue to be $ 1,200?

Answer 1

px = 1200

p(800 - 100p) = 1200

800p - 100p² = 1200

-100p² + 800p - 1200 = 0

-100(p² - 8p + 12) = 0

-100(p -2)(p - 6) = 0

 

The price could be either $2 or $6.

Answer 2

daily revenue = xp = 1200, x= 1200/p

 

also x = 800 - 100p

 

plug in x = 1200/p in this equation

 

1200/p = 800 - 100p

 

1200 = 800p - 100p²

 

100p² - 800p +1200 = 0

 

solve this quadratic equation for p

 

100p² - 600p -200p +1200 = 0

 

100p(p-6) - 200(p-6) = 0

(100p - 200)(p-6) = 0

 

p=2 or p=6

:)