Answer:
The dimension of box=[tex]2ft\times 2ft\times 1ft[/tex]
Step-by-step explanation:
We are given that
Volume of box=4 cubic feet
Let x be the side of square base and h be the height of box
Volume of box=[tex]lbh=x^2h[/tex]
[tex]4=x^2h[/tex]
[tex]h=\frac{4}{x^2}[/tex]
Now, surface area of box,A=[tex]x^2+4xh[/tex]
[tex]A=x^2+4x(\frac{4}{x^2})=x^2+\frac{16}{x}[/tex]
[tex]\frac{dA}{dx}=2x-\frac{16}{x^2}[/tex]
[tex]\frac{dA}{dx}=0[/tex]
[tex]2x-\frac{16}{x^2}=0[/tex]
[tex]2x=\frac{16}{x^2}[/tex]
[tex]x^3=8[/tex]
[tex]x=2[/tex]
[tex]\frac{d^2A}{dx^2}=2+\frac{32}{x^3}[/tex]
Substitute x=2
[tex]\frac{d^2A}{dx^2}=2+\frac{32}{2^3}=2+4=6>0[/tex]
Hence, the area of box is minimum at x=2
Therefore, side of square base,x=2 ft
Height of box,h=[tex]\frac{4}{2^2}=1 ft[/tex]
Hence, the dimension of box=[tex]2ft\times 2ft\times 1ft[/tex]