CABChp4-4Prob27

# CABChp4-4Prob27 - P(n = 6600 68n 2n 2 maximum profit set...

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CalcAB Chp 4-4 Help with Pg 228 #27 Maximize profit Option #1: Instead of setting up the problem using x = # of people Set up a new variable; n, it is used as a counter for the number of changes made to the people and price per person. # people (#units) \$ per person (price per person) n counts the # of changes 50 \$200 0 (no change) 51 \$198 1 52 \$196 2 53 \$194 3 (50+n) (200 2n) revenue = (#units) (price per person) R(n) = (50+n)(200 2n) cost = (cost per unit) (#units) + fixed amount C(n) = 32(50+n) + 6000 profit = revenue cost P(n) = (50+n)(200 2n) [32(50+n) + 6000 ] simplify P(n) = 1000 +100n 2n 2 [ 1600 + 32n + 6000 ]

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Unformatted text preview: P(n) = 6600 + 68n 2n 2 maximum profit set P’(n)=0 P’(n) = 68 4n = n = 17 Question asks how many people will max. profit # people = 50 + 17 = 67 people Option #2: Use x = # of people price per person: 200 2(x 50) (price goes down 2 for each person more than 50) revenue: R(x) = x [ 200 2(x 50) ] simplify: R(x) = 300x 2x 2 cost: C(x) = 6000 + 32x (\$6000 is fixed and each person costs \$32) profit: P(x) = 300x 2x 2 [6000 + 32x] simplify: P(x) = 2x 2 + 268x 6000 derivative: P'(x) = 4x + 268 = 0 x = 67 people will maximize profit...
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## This note was uploaded on 02/21/2011 for the course MATH 408C taught by Professor Knopf during the Spring '10 term at University of Texas.

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CABChp4-4Prob27 - P(n = 6600 68n 2n 2 maximum profit set...

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