Answer
Dimensions $2\times6\times12\ in$, max volume $144\ in^3$
Work Step by Step
Step 1. Assume the cut out squares have an edge length of $x$; the dimensions of the box can be found as height=$x$, width=$10-2x$, and length=$16-2x$ where $0\leq x\leq 5$ to make sure all the dimensions are positive.
Step 2. The volume is given by $V=x(10-2x)(16-2x)=4(x^3-13x^2+40x)$
Step 3. Take the derivative of the function to get $V'=4(3x^2-26x+40)$ and let $V'=0$ to get $(3x-20)(x-2)=0$ and only $x=2$ is in the domain.
Step 4. Check $V''(2)=4(6x-26)=-56\lt0$ and we confirm $x=2\ in$ gives a maximum volume of $144\ in^3$; thus we get all the dimensions as $2\times6\times12\ in$
Step 5. See graph; we can find a maximum of the volume at $(2,144)$.
