Confusing Cuboids

The proof will be posted later. The dimensions of the cuboids are $4 \times 4 \times 1$ and $2 \times 2 \times 5$ , the volumes being $16+20 =36$ , while $E=36$ and $A=48$ .

Edit:

The dimensions $x,y,z$ in the following is interchangeable.

Let $x,y,z$ be the dimensions of a cuboid, and

$E=4\left( x+y+z \right)$
$A=2\left( xy+yz+zx \right)$

We solve for $x,y$ , the solutions being

$x=\dfrac { 1 }{ 8 } \left( E-4z-\sqrt { { E }^{ 2 }-32A+8Ez-48{ z }^{ 2 } } \right)$
$y=\dfrac { 1 }{ 8 } \left( E-4z+\sqrt { { E }^{ 2 }-32A+8Ez-48{ z }^{ 2 } } \right)$

or

$x=\dfrac { 1 }{ 8 } \left( E-4z+\sqrt { { E }^{ 2 }-32A+8Ez-48{ z }^{ 2 } } \right)$
$y=\dfrac { 1 }{ 8 } \left( E-4z-\sqrt { { E }^{ 2 }-32A+8Ez-48{ z }^{ 2 } } \right)$

The volume $xyz$ works out to

$\dfrac{1}{4}\left(2Az-E{z}^{2}+4{z}^{3}\right)$

which is a real value for all real $E,A,z$ . Maximum and minimum volumes occurs when

$z=\dfrac { 1 }{ 12 } \left( E-\sqrt { { E }^{ 2 }-24A } \right)$
$z=\dfrac { 1 }{ 12 } \left( E+\sqrt { { E }^{ 2 }-24A } \right)$

which yields the same volumes when

$z=\dfrac { 1 }{ 12 } \left( E+2\sqrt { { E }^{ 2 }-24A } \right)$
$z=\dfrac { 1 }{ 12 } \left( E-2\sqrt { { E }^{ 2 }-24A } \right)$

which is the condition that the radical in $x,y$ in terms of $E,A,z$ vanish, so that all the dimensions are real. Hence for any given real $E,A$ such that $32A>{ E }^{ 2 }>24A$ , the dimensions of the cuboids of maximum and minimum volumes are (in any order of $x,y,z$ )

$\dfrac { 1 }{ 12 } \left( E+2\sqrt { { E }^{ 2 }-24A } \right)$
$\dfrac { 1 }{ 12 } \left( E-\sqrt { { E }^{ 2 }-24A } \right)$
$\dfrac { 1 }{ 12 } \left( E-\sqrt { { E }^{ 2 }-24A } \right)$

and

$\dfrac { 1 }{ 12 } \left( E-2\sqrt { { E }^{ 2 }-24A } \right)$
$\dfrac { 1 }{ 12 } \left( E+\sqrt { { E }^{ 2 }-24A } \right)$
$\dfrac { 1 }{ 12 } \left( E+\sqrt { { E }^{ 2 }-24A } \right)$

For all these to be integers, $E$ must at least be a multiple of $12$ , or $12e$ , and $A$ must at least be a multiple of $6$ , or $6a$ . We try different integer values for $e, a=1,2,3,...$ and the smallest dimensions and volumes possible occur where $e=3$ and $a=8$ , the dimensions being $2 \times 2 \times 5=20$ and $1 \times 4 \times 4=16$ .