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Revenue Practice Problems (page 2)

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By — McGraw-Hill Professional
Updated on Sep 27, 2011

Solutions

Revenue Solutions

If x = 1, the rent should be 400 + 25(1) = $425. If x = 8, the rent should be 400 + 25(8) = $600.

Revenue Solutions

Revenue Solutions

If x = 2.50, the price should be 2.80 – 0.10(2.5) = $2.55. If x = 5.5, the price should be 2.80 – 0.10(5.50) = $2.25.

  1. P = 4 – 0.05 x Q = 500 + 20 x R = (4 – 0.05 x )(500 + 20 x )

2450 = (4 – 0.05 x )(500 + 20 x )

2450 = 2000 + 55 x – x 2

x 2 – 55 x + 450 = 0

( x – 45)( x − 10) = 0

Revenue Solutions

If x = 45, the price should be 4 – 0.05(45) = $1.75 and 500 + 20(45) = 1400 buckets would be sold. If x = 10, the price should be 4 – 0.05(10) = $3.50 and 500 + 20(10) = 700 buckets would be sold.

Revenue Solutions

The price should be 30 + 2(1) = $32. There would be 40 – 5(1) = 35 oil changes performed each day.

  1. P = 1.50 – 0.05 x Q = 250 + 10 x

R = (1.50 – 0.05 x )(250 + 10 x )

378 = (1.50 – 0.05 x )(250 + 10 x )

378 = 375 + 2.5 x – 0.5 x 2

0.5 x 2 – 2.5 x + 3 = 0

2(0.5 x 2 – 2.5 x + 3) = 2(0)

x 2 – 5 x + 6 = 0

( x – 2)( x – 3) = 0

Revenue Solutions

If x = 2, the price should be 1.50 – 0.05(2) = $1.40. If x = 3, the price should be 1.50 – 0.05(3) = $1.35.

Revenue Solutions

( x = 56 is not likely to be a solution—the price would be $152!) If the price of the shoes are 40 + 2(4) = $48 per pair, the revenue will be $3648.

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