Let x ft be the length of the base square and y ft be the height of the box.
The volume of the box is
[tex]V=x\cdot x\cdot y\ ft^3.[/tex]
Since the box has a volume of 15 cubic feet, then
[tex]x^2y=15,\\ \\y=\dfrac{15}{x^2}.[/tex]
You need to construct two squares from the metal that cost $3 per square foot.
The area of each square is [tex]x^2\ ft^2[/tex] and the total cost for these two squares is [tex]2\cdot x^2\cdot 3=6x^2.[/tex]
The area of each side face is [tex]x\cdot y=x\cdot \dfrac{15}{x^2}=\dfrac{15}{x}.[/tex] Then the total cost for sides is [tex]4\cdot \dfrac{15}{x}\cdot 10=\dfrac{600}{x}.[/tex]
Let S(x) be the function that represents total cost of the box, then
[tex]S(x)=6x^2+\dfrac{600}{x}.[/tex]
Find the derivative:
[tex]S'(x)=12x-\dfrac{600}{x^2}.[/tex]
When [tex]S'(x)=0,[/tex] then [tex]12x-\dfrac{600}{x^2}=0,\\ \\12x^3=600,\\ \\x^3=50,\\ \\x=\sqrt[3]{50}\ ft.[/tex]
The dimensions of the box are:
length and width - [tex]\sqrt[3]{50}\ ft[/tex]
height - [tex]\dfrac{15}{\sqrt[3]{2500}}\ ft.[/tex]
