Answer:
[tex]512\ \text{in}^2[/tex]
Step-by-step explanation:
x = Length and width of base
y = Height of box
Volume of the box is [tex]16384\ \text{in}^3[/tex]
[tex]x^2y=16384\\\Rightarrow y=\dfrac{16384}{x^2}[/tex]
Surface area is given by
[tex]s=x^2+4y\\\Rightarrow s=x^2+4\times \dfrac{16384}{x^2}\\\Rightarrow s=x^2+\dfrac{65536}{x^2}[/tex]
Differentiating with respect to x we get
[tex]s'=2x-\dfrac{131072}{x^3}[/tex]
Equating with 0 we get
[tex]0=2x^4-131072\\\Rightarrow x=(\dfrac{131072}{2})^{\dfrac{1}{4}}\\\Rightarrow x=16[/tex]
[tex]s''=2+\dfrac{393216}{x^4}[/tex]
at [tex]x=16[/tex]
[tex]s''=2+\dfrac{393216}{16^4}=8>0[/tex]
So the function is minimum at x = 16
[tex]y=\dfrac{16384}{x^2}=\dfrac{16384}{16^2}\\\Rightarrow y=64[/tex]
The material required is
[tex]s=x^2+4y=16^2+4\times 64\\\Rightarrow s=512\ \text{in}^2[/tex]
The minimum amount of material required is [tex]512\ \text{in}^2[/tex].
