Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.1 - Real Numbers - 1.1 Exercises - Page 12: 87

Answer

Part A: Package that is 6 in. wide, 8 in. deep, and 5 ft long would be accepted. Package that measures 2 ft by 2 ft by 4 ft would not be accepted. Part B: The greatest acceptable length for a package that is 9 inches wide and 9 inches deep is 72 inches.

Work Step by Step

Given formula: L + 2(x+y) $\leq$ 108 inches Things to keep in mind: L represents the length (how long it is) x and y represent the girth (how wide and deep it is) Our unit is inches, so at times we will convert from feet to inches. Part A: Will a package 6 inches wide, 8 inches deep, and 5 feet long be accepted? To be accepted the measurements must $\leq$ 108 inches. 6 inches wide and 8 inches deep represent x and y, 5 feet long represents L. Since our unit is inches, we have to covert the 5 feet to inches. Conversion: 1 foot equals 12 inches, so 5 feet is 60 inches. Putting everything into our given formula yields: 60 + 2 (6 + 8) $\leq$ 108 ? 60 + 2 (14) $\leq$ 108 ? 60 + 28 $\leq$ 108 ? 88 $\leq$ 108 ✓ Therefore, a package with these dimensions would be accepted. Will a package 2 feet wide, 2 feet deep, and 4 feet long be accepted? We must convert from feet to inches: 2 feet is 24 inches, and 4 feet is 48 inches Putting this into the formula yields: 48 + 2 (24 + 24) $\leq$ 108 ? 48 + 2 (48) $\leq$ 108 ? 48 + 96 $\leq$ 108 ? 144 $\leq$ 108 ✗ This is false because 144 is greater than 108. Therefore, a package with these dimensions would not be accepted. Part B: What is the greatest acceptable length for a package that is 9 inches wide, and 9 inches deep? Putting the known dimensions into our formula yields: L + 2 (9+9) $\leq$ 108 L + 2(18) $\leq$ 108 L + 36 $\leq$ 108 L $\leq$ 72 Therefore, the length must be less than or equal to 72 inches. In other words, the greatest acceptable length for a package with the given dimensions is 72 inches.
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