The question is worded as follows:
"A courier firm has restrictions on the size of parcels that it will carry at the normal rate. The length of a parcel must be no less than twice its width. The sum of the length, width and height of any parcel also must be no more than 136 millimetres."
I thought I had solved this problem, I wrote an equation for the Volume, as well as a function graphing the width. Turns out I was way off and totally messed it up. I just can't understand how to solve this as the only way I can express $l$ seems to be with an inequality and I can't solve that. When I solved it first I tried: $l = 2w - x$
But I don't think that's correct. Any help would be appreciated. I'm also asked to provide an assumption which is made about the shape of the package. Thanks.
Calling
$$ x = \mbox{length}\\ y = \mbox{width}\\ z = \mbox{height} $$ the restrictions can be established as
$$ x \ge 0\\ y \ge 0\\ z \ge 0\\ x\ge 2 y\\ x+y + z \le 136 $$
resulting in the restriction volume attached
so the maximal feasible dimensions volume in one piece is
$$ x = \frac{544}{9}\\ y = \frac{272}{9}\\ z = \frac{136}{3} $$
Assuming that the maximum volume fits tight we can proceed as follows
$$ V = x y z\\ x = 2y\Rightarrow V = 2y^2z\\ x+y+z = L\Rightarrow 3y + z = L\Rightarrow z = L - 3y\\ V = 2y^2(L-3y) $$
so the extremum points obey
$$ \frac{d}{dy}V(y) = 4yL-18y^2=0\Rightarrow y = \frac{272}{9} $$
etc.