Asked by Brendan Zankowski
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.