Answer
$V=\frac{2450}{27}\approx90.74in^3$, dimensions $\frac{5}{3}\times\frac{14}{3}\times\frac{35}{3}$in.
Work Step by Step
Step 1. Drawing a figure as shown, assume the side length of the cut out square is $x$ in with $0\leq x\leq 4$.
Step 2. We can identify the dimensions of the box as height=$x$, length=$15-2x$, width=$8-2x$
Step 3. The volume of the box is given by $V=x(8-2x)(15-2x)=2(x^2-4x)(2x-15)$
Step 4. Take the derivative to get $V'=2(2x-4)(2x-15)+2(x^2-4x)(2)=4(x-2)(2x-15)+4(x^2-4x)=4(2x^2-19x+30+x^2-4x)=4(3x^2-23x+30)=4(3x-5)(x-6)$
Step 5. The extreme of the volume happens when $V'=0$ which gives $x=\frac{5}{3},6$. Discard $x=6$ as it is not in the domain. Thus we have $x=\frac{5}{3}$ and $V=\frac{5}{3}(8-2(\frac{5}{3}))(15-2(\frac{5}{3}))=\frac{5(24-10)(45-10)}{27}=\frac{2450}{27}\approx90.74in^3$
Step 6. We can determine the above value is a maximum by testing signs of $V'$, as $V'(0)=120, V'(4)=-56$, we have $(0)..(+)..(\frac{5}{3})..(-)..(4)$ which means we have a maximum at $x=\frac{5}{3}$ with $V=\frac{2450}{27}\approx90.74in^3$ and dimensions $\frac{5}{3}\times\frac{14}{3}\times\frac{35}{3}$in.
