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?

the volume would be V₂ = 5225 in³

Explanation:

denoting w as width, L as length and h as height, then the volume of the box will be

V = w*h*L

the surface area will be

S = 2*w*h + 2*w*L + 2*h*L→ S/2 = w*h + w*L + h*L

and the sum of the lengths TL will be

TL = 4*h + 4*L + 4*w → TL/4 = w + h + L

if the dimensions would increase 1 unit, the volume would be

V₂ = (w+1) * (h+1) * (L+1) = w*h*L + w*h + wL + w + hL + h + L + 1 = w*h*L + (w*h + wL+hL) + (w+h+L) + 1 = V + S/2 + TL/4 + 1

V₂ = V + S/2 + TL/4 + 1 = 4320 in³ + 1704 in²/2 * 1 in + 208 in/4 * 1 in² + 1 * 1 in³ = 5225 in³

V₂ = 5225 in³