The box should be a cube and each dimension should be \frac{2\sqrt{21}}{3}\approx 3.05 units. The maximum volume is \frac{56\sqrt{21}}{9}\approx 28.5 cubic units.
Let x, y, and z be the dimensions of the box. Then V=xyz and S=2xy+2xz+2yz are the volume and surface areas of the box.
It might make some intuitive sense to you that the answer is a cube so that x=y=z. If it does, then the surface area being 56 square units implies that S=6x^2=56 so that x^2=y^2=z^2=\frac{56}{6}=\frac{28}{3} and x=y=z=\sqrt{\frac{28}{3}}=\frac{2\sqrt{7}}{\sqrt{3}}=\frac{2\sqrt{21}}{3} and V=xyz=(\frac{2\sqrt{21}}{3})^3=\frac{8\cdot 21\sqrt{21}}{27}=\frac{56\sqrt{21}}{9}
More rigorously, we can use calculus:
Since S=56, we can say that 2z(x+y)=56-2xy so that z=\frac{28-xy}{x+y}. This means we can write the volume as a function of two variables:
V=\frac{xy(28-xy)}{x+y}=\frac{28xy-x^2y^2}{x+y}.
The partial derivatives of this function are:
\frac{\partial V}{\partial x}=\frac{(x+y)(28y-2xy^2)-(28xy-x^2y^2)\cdot 1}{(x+y)^2}=\frac{y^2(28-x^2-2xy)}{(x+y)^2}
and
\frac{\partial V}{\partial y}=\frac{(x+y)(28x-2x^2y)-(28xy-x^2y^2)\cdot 1}{(x+y)^2}=\frac{x^2(28-y^2-2xy)}{(x+y)^2}.
Setting these both equal to zero results in the system of equations x^2+2xy=28, y^2+2xy=28. The first of these equations can be solved for y to get y=\frac{28-x^2}{2x}.
Plugging this into the second equation leads to (\frac{28-x^2}{2x})^{2}+2x\cdot \frac{28-x^{2}}{2x}=28. Multiplying both sides of this last equation by 4x^2 gives (28-x^2)^2+112x^2-4x^4=112x^2, which reduces to 3x^4+56x^2-784=0.
Using the quadratic formula, x^2=\frac{-56\pm\sqrt{56^2-4\cdot 3\cdot (-784)}}{2}=\frac{-56\pm\sqrt{12544}}{6}=\frac{-56\pm 112}{6}. The positive one is x^2=\frac{56}{6}=\frac{28}{3} so that x=\sqrt{\frac{28}{3}}=\frac{2\sqrt{7}}{\sqrt{3}}=\frac{2\sqrt{21}}{3}\approx 3.05. The symmetry in the original equations implies that y=\frac{2\sqrt{21}}{3}\approx 3.05 as well. If you use the equation z=\frac{28-xy}{x+y} you'll get z=\frac{2\sqrt{21}}{3}\approx 3.05 as well. That this gives a maximum value can be checked with the 2nd derivative test.
