Question 198164
Let {{{w}}}= the width in inches
{{{w + 2}}}= the length
{{{w - 1}}}= the height
Volume = length x width x height
{{{w*(w + 2)*(w - 1) = 140}}} in3
{{{w^2 + 2w)*(w - 1) = 140}}} in3
{{{w^3 + 2w^2 - w^2 - 2w = 140}}}
{{{w^3 + w^2 - 2w = 140}}}
factor out {{{w}}}
{{{w*(w^2 + w - 2) = 140}}}
Factoring again:
{{{w*(w + 2)(w - 1) = 140}}}
I'll say {{{y = w*(w + 2)(w - 1)}}}
I'll plot it and see where {{{y = 140}}} is
{{{ graph( 400, 400, -5, 5, -5, 50, x^3 + x^2 - 2x) }}}
It looks like {{{w}}} is going to be about {{{4}}}
{{{y(4) = 4*(4 + 2)*(4 - 1)}}}
{{{y(4) = 4*6*3}}}
{{{y(4) = 72}}}
I'll try {{{5}}}
{{{y(5) = 5*7*4}}}
{{{y(5) = 140}}} bingo
The width is 5 inches