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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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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

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