From the figure, \(l=3x−5 ft, w=2x−1 ft,\) and \(h=2 ft\). The volume is \(40 ft^3\) so we can write:

\(V=lwh\)

\(40=(3x−5)(2x−1)(2)\)

Divide both sides by 2:

\(20=(3x−5)(2x−1)\)

Expand the right side:

\(\displaystyle{20}={6}{x}^{{2}}−{13}{x}+{5}\)

Write in standard form:

\(\displaystyle{0}={6}{x}^{{2}}−{13}{x}−{15}\)

Factor the right side:

\(0=(6x+5)(x−3)\)

By zero-product property,

\(x=−56,3\)

Since \(x=−56\) gives negative length and width, then it is extraneous. Hence,

\(x=3\)

Using the value of \(xx\), the length is \(3(3)−5=4\) ft ft and the width is \(2(3)−1=5 ft.\)