Your college newspaper, The Collegiate Investigator, sells for 90c per copy. The

Question

Your college newspaper, The Collegiate Investigator, sells for 90c per copy. The cost of producing x copies of an edition is given by C(x) = 30 + 0.10x + 0.001x^2 dollars. (a) Calculate the marginal revenue R'(x) and profit P'(x) functions. R'(x) = __________ P'(x) = ___________ (b) Compute the revenue and profit, and also the marginal revenue and profit, if you have produced and sold 500 copies of the latest edition, revenue $ __________ profit $ __________ marginal revenue $ ____________ per additional copy marginal profit $ ______ per additional copy Interpret the results. The approximate from the sale of the 501st copy is $ _________. (c) For which value of x is the marginal profit zero? x = ____________ copies Interpret your answer. The graph of the profit function is a parabola with a vertex at x = _______, so the profit is at a maximum when you produce and sell ________ copies.

Explanation / Answer

We have given C(x)=30+0.10x+0.001x^2 dollars and sells for 90¢ per copy=0.90 dollars

a) Revenue R(x) =Price*Quantity =0.90x dollars

Profit P(x)=R(x)-C(x) =0.90x-(30+0.10x+0.001x^2)

R'(x)=0.90

P'(x)=0.90-0.10-(0.002)x=0.80-(0.002)x

P'(x)=0.80-(0.002)x

b) Revenue R(x) =Price*Quantity =0.90*500=$450

Profit P(x)=R(x)-C(x)=450-(30+(0.10*500)+(0.001*(500)^2))=$120

Marginal Revenue is R'(x)=$0.90 per additional copy

Marginal Profit P'(x)=0.80-(0.002)*500=-$0.2 per additional copy

interpret the results

R'(x)=$0.90

R'(500)=$0.90

the approximate revenue from the sale of the 501st copy is $0.90

c) the marginal profit zero when

P'(x)=0.80-(0.002)x=0

0.002x=0.80

x=0.80/0.002=400

x=400 copies the marginal profit zero

when selling more than 400 copies,the additional profit per copy is 0

Profit function P(x)= 0.90x-(30+0.10x+0.001x^2)

P(x)=-0.001x^2+0.80x-30

comparing with parabola y=ax^2+bx+c

y=-0.001x^2+0.80x-30

a=-0.001,b=0.80 and c=-30

x-coordinate of vertex is -b/2a

=-0.80/2*(-0.001)=400

the graph of the profit function is a parabola with a vertex at x=400

so substitute x=400 into P(x)=-0.001x^2+0.80x-30

y=(-0.001)*(400)^2+0.80*(400)-30=130

so the profit is at a maximum when you produce and sell 400 copies

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