A rectangular box has a volume of $4320$ cubic inches and a surface area of $1704$ square inches. The sum of the lengths of its $12$ edges is $208$ inches. What would be the volume of the box, in cubic inches, if its length, width and height were each increased by one inch?

Answer 1

looks like we have 3 unknowns :
l, w, and h for length , width and height

given: lwh = 4320 **
2lw + 2lh + 2wh = 1704 ---> lw + lh + wh = 852 ***
4l + 4w + 4h = 208 ----> l+w+h = 52 ****
from ****, l = 52-w-h
from ** , l(w+h) + wh = 852 ----> l = (852-wh)/(w+h)

52-w-h = (852-wh)/(w+h)
852-wh = 52w + 52h - w^2-wh -wh - h^2
w^2 + h^2 - 52w + wh - 52h + 852 = 0 obtained only from *** and ****
have not involved ** yet

from ** , l = 4320/(wh)
combine that with l = 52-w-h to get
4320/(wh) = 52-w-h

arghhh, getting too messy to type.
I will give you a hint: w = 18
from 4320/(wh) = 52-w-h, you can now get h
then go back to lwh = 4320 to find l