Question 706282
x length, y width, z height.
volume of the box would be xyz.

x=2y as given, so volume is 2*y*y*z or volume {{{v=2y^2z.}}}.


Surface area was given as 300 cubic units.   Accounting for the lengths measurements,
{{{2*2*y*y+2yz+2*y*2*z=300}}}
4y^2+2yz+4yz=300
...and few more steps...
z=(300-4y^2)/(2y+4y)
{{{z=(2/3)(75-y^2)/y}}}


Back to the volume formula,
{{{v=2y^2(z)}}}
v=2y^2(2/3)(75-y^2)/y
v=...
{{{v=100y-(4/3)y^3}}} ----Taking derivative of formula in this form may be easiest.


The task to do now is the maximize v as a function of width y.   Differentiate v with respect to y.
You then simply calculate x from knowing how x and y were given.  Then you may need to use the z formula found above to caclulate z.