Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises - Page 64: 79


The dimensions of the box: - Height: $6$ ft - Length: $15$ ft - Width: $2$ ft

Work Step by Step

$Volume = 180$, $x>0$ $x(x-4)(x+9)=180$ $x(x^{2}+9x-4x-36)=180$ $x(x^{2}+5x-36)=180$ $x^{3}+5x^{2}-36x-180=0$ $x^{2}(x+5)-36(x+5)=0$ $(x+5)(x^{2}-36)=0$ $x+5=0$ or $x^{2}-36=0$ With $x+5=0$, so $x=-5$ With $x^{2}-36=0$, so $x=6$ or $x=-6$ Because $x>0$, so $x=6$ Therefore, the dimensions of the box: - $Height=x=6$ ft - $Length=x+9=6+9=15$ ft - $Width=x-4=6-4=2$ ft
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