Arectangular 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

the volume would be V₂ = 5225 in³

Explanation:

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