MATH HOMEWORK !

Question

1. The demand for bricks depends on the price. A brick company can sell 300 000 bricks at $4.00 each. For every $0.20 the price is raised, they sell 10 000 fewer bricks. In addition, the company’s cost, C in dollars, of producing each brick is related to the number of bricks to be sold, n, where 𝐶 = 3. 5 - 𝑛/1000000

a. Create an equation to represent the company's profit, P, in terms of x, the number (Note: Profit will equal revenue−cost.)

b. Find the selling price (to the nearest cent) and number of bricks sold that will lead to a profit of $750 000

Answer 1

C= 3.5 - n/1,000,000

Profit = P= R-C = Revenue minus Cost

R = np where p = price = -n/50,000 + 10

R = -n^2/50000 +10n

P =- n^2/50000 +10n - 3.5 + n/1,000,000

P =- n^2/50000 + 10.000001n -3.5

or use x for n

sales for $750,000 profit is

the x solution of this equation

750,000= -x^2/50000 + 10.000001x -3.5

use the quadratic equation

x = 10.000001/1/50000 +/- (1/50000)sqr(10.000001^2 -(4/50000)(-750,003.5))

300,000 sold at $4, 10,000 less for every price increase of 20 cents

(n,p) =

(300,000, 4), (290,000, 4.2) are two points with slope = -.2/10,000 =

-1/50,000 = (p-4)/(n-300,000)

50,000p -200,000 = -n +300,000

50000p = -n +500,000

p = -n/50000 + 10

to find the price that gives $750,000 profit plug the x value found with the quadratic equation above and substitute it for n in p=-n/50000+10