Question

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?

Answer 1

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

Answer 2

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