Answer
a. $h=24in$, $w=18in$ and $V=10368in^3$
b. $V(24)=10368in^3$, see graph.
Work Step by Step
a. Step 1. In this case, use the given figure; let $h+2h+2w=108$ or $w=\frac{108-3h}{2}$ as required by the U.S. Postal Services and the volume is then given by $V=h^2w=h^2\cdot \frac{108-3h}{2}=\frac{108h^2-3h^3}{2}$
Step 2. Take the derivative to get $V'=\frac{216h-9h^2}{2}$. Letting $V'=0$, we get $h(216-9h)=0$. Discard zero and negative solutions; we have $h=24in$, which gives $w=18in$ and $V=24^2\times18=10368in^3$
Step 3. Check $V''=\frac{216-18h}{2}|_{h=24}=-108\lt0$; thus the function is concave down at this point and the volume is a maximum.
b. Graph the function $V=\frac{108h^2-3h^3}{2}$ as shown. We can find a maximum at $h=24$ with $V(24)=10368in^3$ which agrees with the results from part (a).
