The cost to produce one laptop computer is 890.00 plus a one time fixed cost of 100,000.00 for research and development. Each laptop computer will be sold for 1520.00.

A. Write a formula that gives the cost C of producing x laptops

B. Write a formula that gives the revenue R from selling x laptop computers

C. Profit equally revenue minus cost. Write formula that calculates the profit P from selling X laptop computers.

D. Set P>0 and solve for x. Interpret your answer

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

a)

C = 890 X + 100,000

b)

R = 1520 X

c) P= 1520X - 890X - 100,000

d> 1520 X -890X - 100,000 >0

630 X > 100,000

X > 100,000/630 = 158.7

Has to sell at least 159 laptop to make a profit.

159 sold is a breakeven point.

Dear Jason,

(A.) The cost, C, of producing x laptops would be

C = ($890)*x + ($100,000)/x

(B.) Revenue, R, from selling x laptops would be

R = ($1520)*x

(C.) The profit, P, from selling x laptops is

P = ($1520)*x - [($890)*x + ($100,000/x)] or ($630)*x - ($100,000/x)

(D.) Let P = 0

Then

($630)*x - ($100,000/x) = 0

($630)*x = ($100,000/x)

Multiply both sides by x

($630)*x^{2} = $100,000

Divide both sides by $630

x^{2} = 158.73

x = 12.5988

In this case x is the minimum number of laptops that can be made and sold to start realizing a profit. However, since no one is going to make, let alone sell, 0.5988 of a computer, the company will have to make at least 13 laptops to start realizing a profit.

A. C = 100000 + 890 x

B. R = 1520 x

C. Profit = Revenue - Cost; P = R - C = 1520 x - 890 x - 100000 = 630 x - 100000

D. P > 0 => 630 x - 100000 > 0 => x > 100000/630 ≈ 158.73015873015873,

but since x has to be an integer, x > 158 for a positive profit. If x <= 158 there will be a loss.

D) P = 1520x - (890x + 100,000)

= 1520x - 890x - 100,000

P = 630x - 100,000

P + 100,000 = 630x

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