Répondre :
if x >15 then the box can't exist !!
V(x)=(30-2x)²x=(900-120x+4x²)x=4x^3-120x²+900x
V'(x)=12x²-240x+900 =12(x²-20x+75) =12((x-10)²-25)=12(x-10+5)(x-10-5)=12(x-5)(x-15)
Then V is increasing between 0 and 5 and decreasing on 5,15
So the maximum of V is V(5) also 400*5=2000
this kind of box has the dimensions 20, 20, and 5