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?

Answers

Answered by Reiny
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

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