Answer
(a) see prove below.
(b) $1.45ft\times1.45ft\times1.34ft$ and $2.31ft\times2.31ft\times0.53ft$
Work Step by Step
(a) Assume the height is $y$, the known diagonal can be expressed as
$ {y^2+(\sqrt 2x)^2}=(x+1)^2$ (Pythagorean Theorem for the right triangle involving the diagonal)
so $y=\sqrt {(x+1)^2-2x^2}=\sqrt {1+2x-x^2}$ the known volume can then be expressed as
$x^2y=x^2\sqrt {1-+2x-x^2}=2\sqrt 2$ transform this equation as
$x^4(1+2x-x^2)=8$ which gives $x^6-2x^5-x^4+8=0$
(b) Graph the function and we can find two solutions as $x=1.45,2.31ft$ for two cases,
and the dimensions are $1.45ft\times1.45ft\times1.34ft$ and $2.31ft\times2.31ft\times0.53ft$
