College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 140: 86

Answer

Length = 2 in Width = 15 in Height = 6 in

Work Step by Step

Let $x$ ft be the height of the box. Based on the given conditions, we have: $Volume=length\times width\times height$ $180=(x-4)x(x+9)$ Solve for $x$: $(x-4)x(x+9)=180$ $x^3+5x^2-36x=180$ $x^3+5x^2-36x-180=0$ (Factorize) $(x+6)(x+5)(x-6)=0$ $x=-6$ or $x=-5$ or $x=6$ Since $x$ must be positive, the solution is $x=6$. Therefore, (i) The length is $x-4=6-4=2$ in (ii) The width is $x+9=6+9=15$ in (iii) The height is $x=6$ in
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