Application of Derivatives Part 2

Length of box = a − 2x
Breadth of box = b − 2x
Height of box = x
Volume of box, V = (a − 2x) (b − 2x)x
⇒V=4x3−20x2−2bx2+abx
Differentiating V w.r.t. x,
V′(x)=12x2−4(a+b)x+ab
Critical Point, V′(x)=0
⇒12x2−4(a+b)x+ab=0
⇒x=

4(a+b)±√16(a+b)2−4.12.ab

2(12)

x=

(a+b)±√a2+b2−ab

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V′′(x)=24x−4(a+b)
Forx=

(a+b)−√a2+b2−ab

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, V′ ′(x) < 0
Hence, for maximum volume
x=

a+b−√a2+b2−ab

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