Math question:An open-top box with a square base is to have a volume of 4 cubic ft. Find the dimensions of the box that can be be made with the smallest amount of material.
This is the only thing I got: $V=x^2y=4$
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$\begingroup$So you have $V=x\cdot x \cdot y=4, $ i.e. your volume is $4$ and this equation has two variables. We need a second equation that way we will have $2$ equations with $2$ variables right?
So whats the surface area of an open top box? In general the it will be (using $w$=width for base, $l$=length for base, and $h$=height) $S=w\cdot l+2 \cdot w \cdot h+2\cdot l\cdot h $ (only one $w \cdot l$ because its open top). Based on your question it looks like $y$ is the height and $w=l=x$.
Solve $V$ for one variable say $y=\frac{4}{x^2}$. Put that into $S$ and calculate $\frac{dS}{dy}=0$. Find the $y$ works and if there's only one its probably the min. If there's more than one you have to check to the right and left to see if its a min or max.
$\endgroup$ $\begingroup$The smallest area:volume ratio (or the largest volume:area) (rectangluar) is found in a cube, i.e. when $x=y=z$ in $3$ dimensions. So, you have $x^2y=4=x^3\implies x=y=\sqrt[3]{4}.$
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