Answer to Question #286100 in Calculus for lux

Question

A delivery company accepts only rectangular boxes the sum of whose length and the perimeter of a cross-section does not exceed 108 inches. Find the dimensions of an acceptable box of largest volume. 

Answer 1

L+4S=108

L=108-4S

$v=s^2L\\ v=s^2(108-4s)\\ v=108s^2-4s^3\\ v'=216s-12s^2\\ 0=216s-12s^2\\ s=0,s=18\\ then, L=108-4(18)\\ L=36.\\ v=s^2L, v= 18^2(36),\\ v=11664in^3$