Stitz-Zeager_College_Algebra_e-book

# A couple of things about the last example are worth

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Unformatted text preview: 1 2 3 4 5 x x 4 (f) g (x) = x(x + 2)3 x = 0 multiplicity 1 x = −2 multiplicity 3 (d) Z (b) = b(42 − b2 ) √ b = − 42 multiplicity 1 b = √multiplicity 1 0 b = 42 multiplicity 1 y Z −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 b −2 −1 x 3. (a) Our ultimate goal is to maximize the volume, so we’ll start with the maximum Length + Girth of 130. This means the length is 130 − 4x. The volume of a rectangular box is always length × width × height so we get V (x) = x2 (130 − 4x) = −4x3 + 130x2 . (b) Graphing y = V (x) on [0, 33] × [0, 21000] shows a maximum at (21.67, 20342.59) so the dimensions of the box with maximum volume are 21.67in. × 21.67in. × 43.32in. for a volume of 20342.59in.3 . (c) If we start with Length + Girth = 108 then the length is 108 − 4x and the volume is V (x) = −4x3 + 108x2 . Graphing y = V (x) on [0, 27] × [0, 11700] shows a maximum at (18.00, 11664.00) so the dimensions of the box with maximum volume are 18.00in. × 18.00in. × 36in. for a volume of 11664.00in.3 . (Calculus will conﬁrm that the measurements which maxi...
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## This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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