Answer:
2m x 2m x 1m
Step-by-step explanation:
Let 'L' be the length of each side of the base, and 'H' be the height of the box.
The area and volume of the box are given by:
[tex]A=12=4LH+L^2\\V=HL^2[/tex]
Rewriting 'V' as a function of 'L':
[tex]12=4LH+L^2\\H=\frac{12-L^2}{4L} \\V=(\frac{12-L^2}{4L})L^2\\V=3L-\frac{L^3}{4}[/tex]
The value of 'L' for which the derivate of the volume function is zero is the length that yields the maximum volume:
[tex]\frac{dV}{dL}=0=3-\frac{3L^2}{4}\\L=\sqrt{4}=2\ m[/tex]
For L= 2m, the height 'H' is:
[tex]12=4H*2+2^2\\H=1\ m[/tex]
Therefore, the dimensions of the box should be 2m x 2m x 1m.
