Q. 31

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

Answer :

Given,

• Side of the square piece is 18 cms.

• the volume of the formed box is maximum.

Let us consider,

• ‘x’ be the length and breadth of the piece cut from each vertex of the piece.

• Side of the box now will be (18-2x)

• The height of the new formed box will also be ‘x’.

Let the volume of the newly formed box is :

V = (18-2x)^{2} × (x)

V = (324+ 4x^{2} – 72x) x

V = 4x^{3} -72x^{2} +324x ------ (1)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. __This is because if the function V(x) has a maximum/minimum at a point c then V’(c) = 0.__

Differentiating the equation (1) with respect to x:

-------- (2)

[Since ]

To find the critical point, we need to equate equation (2) to zero.

x^{2} – 12x + 27 = 0

x = 9 or x =3

x= 2

[as x = 9 is not a possibility, because 18-2x = 18-18= 0]

__Now to check if this critical point will determine the maximum area of the box, we need to check with second differential which needs to be negative.__

Consider differentiating the equation (3) with x:

----- (4)

[Since ]

Now let us find the value of

As , so the function V is maximum at x = 3cm

Now substituting x = 3 in 18 – 2x, the side of the considered box:

Side = 18-2x = 18 - 2(3) = 18-6= 12cm

Therefore side of wanted box is 12cms and height of the box is 3cms.

Now, the volume of the box is

V = (12)^{2} × 3 = 144 × 3 = 432cm^{3}

Hence maximum volume of the box formed by cutting the given 18cms sheet is 432cm^{3} with 12cms side and 3cms height.

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