...haha, "what is the height x, which produces the box with the maximum volume"
V=x(8-2x)(10-2x) (note here that x<4 and x>0 for real solutions)
V=x(4x^2-36x+80)
V=4x^3-36x^2+80x
dV/dx=12x^2-72x+80
d2V/dx2=24x-72
The maximum volume will occur when the velocity dV/dx=0 and the acceleration d2V/dx2<0.
dV/dx=0 when:
12x^2-72x+80=0, using the quadratic formula for expediency:
x=(72±√1344)/24, since 0<x<4
x=(72-√1344)/24 in
x≈1.47 in (to nearest hundredth of an inch)
The maximum volume would be:
V((72-√1344)/24)=4x^3-36x^2+80x
V≈52.51 in^3 (to nearest hundredth of a cubic inch)
