Answer:
See explanation
Step-by-step explanation:
Let x in be the base side length and y in be the height of the box. Since the base is a square, we have
[tex]S=x^2\Rightarrow x=\sqrt{S}[/tex]
The volume of the box is
[tex]V=S\cdot y\\ \\36=Sy\Rightarrow y=\dfrac{36}{S}[/tex]
The surface area of the box is
[tex]SA=2x^2+4xy\\ \\SA(S)=2S+4\cdot \sqrt{S}\cdot \dfrac{36}{S}=2S+\dfrac{144}{\sqrt{S}}[/tex]
The graph of the function SA(S) is shown in attached diagram.
Find the derivative of this function:
[tex]SA'(S)=(2S+144S^{-\frac{1}{2}})'=2-\dfrac{1}{2}\cdot 144\cdot S^{-\frac{1}{2}-1}=2-\dfrac{72}{S\sqrt{S}}[/tex]
Equate this derivative to 0:
[tex]2-\dfrac{72}{S\sqrt{S}}=0\\ \\2S\sqrt{S}=72\\ \\S\sqrt{S}=36\\ \\S^{\frac{3}{2}}=6^2\\ \\S=6^{\frac{4}{3}}[/tex]
So, the dimensions that produce a minimum surface area for this aluminum box are:
[tex]x=\sqrt{6^{\frac{4}{3}}}=6^{\frac{2}{3}} \ in\\ \\y=\dfrac{6^2}{6^{\frac{4}{3}}}=6^{\frac{2}{3}}\ in.[/tex]
