MAA MATHEMATICAL AssociATON OF AMERICA webwork/fal2016.math1120 0030 0/001 week1

Question

MAA MATHEMATICAL AssociATON OF AMERICA webwork/fal2016.math1120 0030 0/001 week13/4 001week13: Problem 4 Problem List sets a week at 510 each Amarket survey indicates that for each 21 rebene oneredtoabuyer the number of sets sold wainorease by 210 per week. How wee rebate should the company ofter toabuyer in oder to the weekly cost functon is 89250 170.how should eset the size of rebate to maximize its profit? Note: can eam partai on this problem. You have attempted this problem0times. You have unlimited attempts nemaining. Logged

Explanation / Answer

a)a point on demand function curve is (1050,510),

after 21 rebate price =510-21=489, number of sets sold =1050+210=1260

another point on curve is (1260,489)

for demand function

p-510=[(489-510)/(1260-1050)](x-1050)

p-510=[-21/210](x-1050)

p-510=-(1/10)(x-1050)

p-510=-(1/10)x+105

p=-(1/10)x+105+510

p=-(1/10)x+615

demand function p(x)=-(1/10)x+615

b) revenue R=p*x

R=(-(1/10)x+615)*x

R=(-(1/10)x2+615x)

for maximum revenue dR/dx =0, d2R/dx2 <0

dR/dx =(-(1/5)x+615), d2R/dx2 =-(1/5) <0

(-(1/5)x+615)=0

x=615*5

x=3075

p=-(1/10)x+615

p=-(1/10)3075+615

p=307.5

rebate =510-307.5

rebate the company should offer for maximum revenue =202.5

c)revenue R=(-(1/10)x2+615x) , cost C=89250+170x

profit P= revenue -cost

P=-(1/10)x2+615x-89250-170x

P=-(1/10)x2+445x-89250

for maximum profit dP/dx =0, d2P/dx2 <0

dP/dx =-(1/5)x+445, d2P/dx2 =-(1/5)<0

-(1/5)x+445=0

x =455*5

x=2275

p=-(1/10)x+615

p=-(1/10)2275+615

p=387.5

rebate =510-387.5

rebate the company should offer for maximum profit =122.5

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