When the surrounding flaps are folded up, the base of the box will have dimensions [tex]11-2x[/tex] by [tex]16-2x[/tex], and the box will have a height of [tex]x[/tex]. So the box has volume, as a function of [tex]x[/tex],
[tex]V(x)=(11-2x)(16-2x)x=176x-54x^2+4x^3[/tex]
I don't know what technology is available to you, but we can determine an exact value for [tex]x[/tex] that maximizes the volume by using calculus.
Differentiating [tex]V[/tex] with respect to [tex]x[/tex] gives
[tex]\dfrac{\mathrm dV}{\mathrm dx}=176-108x+12x^2[/tex]
and setting this equal to 0 gives two critical points at
[tex]x=\dfrac{27\pm\sqrt{201}}6\implies x\approx2.1\text{ or }x\approx6.9[/tex]
For the larger critical point we would get a negative volume, so we ignore that one. Then the largest volume would be about 168.5 cubic in.
