pre calcccccccccc

Question

Suppose that the manufacturer of a gas clothes dryer has found that when the unit price is p dollars, the revenue R (in dollars) is R(p)=−6p2+18,000p.(a) At what prices p is revenue zero?

(b) For what range of prices will revenue exceed $2100,000?

Answer 1

a) 18000p -6p^2 = 0

divide by 6

3000p - p^2 = 0

factor out p

p(3000 - p) = 0

set each factor = 0 then

p = 0 and p = 3,000

b) over what range does revenue exceed 2100??

18000p - 6p^2 = 2100,000

divide by 6

3000p - p^2 = 350,000

complete the square or use the quadratic formula

p^2 - 3000p = -350,000

take half 3000, square it and add it to both sides

p^2 -3000p + 1500^2 = -350,000 + 1500^2

(p-1500)^2 = 350,000 + 2,250,000

take square roots of both sides

p -1500 =+ or -sqr1,900,000=about +/-1378.4

add 1500 to both sides

p = about 121.6 and a negative price (just use 0)

that gives you boundaries for the price range desired

Answer 2

(a) R(p) = - 6p² + 18000p = 0; p = 0 or p = 3000;

(b) - 6p² + 18000p > 2100000, p² - 3000p + 350000 < 0; 121.60 < p < 2878.40