Asked by Zach Bringer
No need to show your work, just give me the solution.
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?
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
Bosnian
w = width
l = length
h = height
Volume:
V = w ∙ l ∙ h = 4320
Surface area:
SA = 2 ( w ∙ l + l ∙ h + h ∙ w ) = 1704
Sum of the lengths:
SL = 4 ( w + l + h ) = 208
System of equations:
w ∙ l ∙ h = 4320
2 ( w ∙ l + l ∙ h + h ∙ w ) = 1704
4 ( w + l + h ) = 208
Solution:
w = 24 in , l = 18 in , h = 10 in
New volume:
Vn = ( w + 1 ) ∙ ( l + 1 ) ∙ ( h + 1 )
Vn = ( 24 + 1 ) ∙ ( 18 + 1 ) ∙ ( 10 + 1 )
Vn = 25 ∙ 19 ∙ 11 = 5225 in³
l = length
h = height
Volume:
V = w ∙ l ∙ h = 4320
Surface area:
SA = 2 ( w ∙ l + l ∙ h + h ∙ w ) = 1704
Sum of the lengths:
SL = 4 ( w + l + h ) = 208
System of equations:
w ∙ l ∙ h = 4320
2 ( w ∙ l + l ∙ h + h ∙ w ) = 1704
4 ( w + l + h ) = 208
Solution:
w = 24 in , l = 18 in , h = 10 in
New volume:
Vn = ( w + 1 ) ∙ ( l + 1 ) ∙ ( h + 1 )
Vn = ( 24 + 1 ) ∙ ( 18 + 1 ) ∙ ( 10 + 1 )
Vn = 25 ∙ 19 ∙ 11 = 5225 in³
Answered by
Zach Bringer
OMG great