# A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box.

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A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find this maximum volume

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selected by Side length of big square is 18 cm

Let the side length of each small square be a.

If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with

Length, L = 18 – 2a

Breadth, B = 18 – 2a and

Height, H = a

Assuming,

Volume of box,

V = LBH = a(18 - 2a)2

Condition for maxima and minima is

$\frac{dV}{da}$ = 0

⇒ (18 - 2a)2 + (a)(- 2)(2)(18 - 2a) = 0

⇒(18 - 2a)[(18 - 2a) - 4a] = 0

⇒ (18 - 2a)[18 – 6a] = 0

⇒ a = 3,9

$\frac{d^2V}{da^2}$ = ( - 2)(18 – 6a) + ( - 6)(18 – 2a)

⇒ $\frac{d^2V}{da^2}$ = 24a - 144

For a = 3,

$\frac{d^2V}{da^2}$ = - 72,

⇒ $\frac{d^2A}{d\theta^2}$ < 0

For a = 9,

$\frac{d^2V}{da^2}$ = 72,

⇒ $\frac{d^2A}{d\theta^2}$ > 0

So for A to maximum,

$\frac{d^2A}{d\theta^2}$ < 0

Hence,

a = 3 will give maximum volume.

And maximum volume,

V = a(18 - 2a)2

= 432 cm3