Answer
a. $18\times18\times36in$, $V=11664in^3$.
b. $V(36)=11664in^3$, see graph.
Work Step by Step
a. Step 1. Assume the square end has an edge of $x$ and the length of the box is $y$; we have $4x+y\leq108$ or $y\leq 108-4x$ as the requirement of the U.S. Postal Service.
Step 2. The volume of the box is given by $V=x^2y\leq x^2(108-4x)=108x^2-4x^3$
Step 3. Take the derivative to get $V'=216x-12x^2$ and let it be zero to get $12x(18-x)=0$ which gives $x=0$ and $x=18$ (discard $x=0$ as it does not make a box).
Step 4. Check $V''=216-24x|_{x=18}=-216\lt0$, thus the function is concave down at this point and the volume is a maximum.
Step 5. At $x=18in$, we have $y\leq 36in$ and $V\leq x^2y=11664in^3$.
b. Using $x=(108-y)/4$, we have $V=x^2y=(\frac{108-y}{4})^2y=\frac{1}{16}(y^3-216y^2+11664y)$.
Graph this function as shown in the figure and we can find its maximum at $V(36)=11664in^3$ which agrees with the results from part (a).
