anonymous
  • anonymous
The length of a rectangular box is one inch greater than its width. The height is three times the length. The diagonal of the box exceeds the height by one inch. Find the volume of the box.
Mathematics
chestercat
  • chestercat
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radar
  • radar
w=width length=w+1 height = 3(w+1) diagonal = 3(w+1)+1 volume = width X length X Height or w X (w+1) X 3(w+1) volume =\[3w ^{3}+5w ^{2}+2w\] That would be just fine if I has some more info. I don't know how to use the "diagonal of the box" information. Maybe some one else can figure it out
anonymous
  • anonymous
Thank you anyway!
anonymous
  • anonymous
A solution using Mathematica with comments is attached.
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radar
  • radar
This may work: The diagonal of the box is:\[\sqrt{(\sqrt{l ^{2}+w ^{2})}+h ^{2}}\] this would become when substituting for length, width, and height in terms of w (width) and simplifying:\[\sqrt{ w ^{2}+w ^{2}+2w +1 + 9w ^{2}+18w+9}\] The diagonal was also given as 3(w+1)+1 we can now write:\[3w+4 = \sqrt{11w ^{2}+20w+10}\] squaring both sides getting\[-2w ^{2}+4w+6=0\]dividing thru by -2 getting\[w ^{2}-2w-3=0\]solving for w (w-3)(w+1)=0 w=3 discarding the negative root. w=3 length = 3+1 = 4 height = 3(w+1)=3(4)=12 volume (3)(4)(12)=144 cubic inches.
radar
  • radar
Great, it worked if the solution provided by Mathematica is correct. Thanks robtobey.

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