## anonymous 4 years ago 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.

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

2. anonymous

Thank you anyway!

3. anonymous

A solution using Mathematica with comments is attached.

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.