TASK: Investigation with square sheets: From a square sheet of paper 20 cm by 20 cm, we can make a box without a lid. We do this by cutting a square from each corner and folding up the flaps.

Q-1-Find a relationship (general rule) between the size of paper(y) and the size of cut(x) that produces the maximum volume?

Q- 2-Test the validity of your general rule by using different values of a, b, and Justify your answer and its degree of accuracy.

Q- 3-Discuss the scope or limitations of the general statement.

Q- 4-Draw a graph Volume (V) and side of square (x) with the suitable scales.

Four questions ,it is not possible to answer in this form ,maxumum 1 question you can ask .

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Let we have sheet of paper of size `yxxy` , we wish to prepare a box ,buy cutting square from each corner of size `x xx x` .Each side of

square reduced to length = y-2x , It is clearl to understand `y-2x>0` .

Thus volume of soformed box will be

`V=(y-2x)^2 xx x`

For maximum volume with fixed m ,we need to find derivative ,and solve the problem.

`(dV)/(dx)=2(y-2x)(-2)x+(y-2x)^2`

`=(y-2x)(-4x+y-2x)`

`=(y-2x)(y-6x)`

for max/min `(dV)/(dx)=0` ,

`(y-2x)(y-6x)=0`

`y-6x=0`

`6x=y`

`x=y/6` ,

`y-2x!=0` ,otherwise box is not possible.

`(d^2V)/(dx^2)=-2(y-6x)+(y-2x)(-6)`

`=-2y+12x-6y+12x`

`=-8y+24x`

`((d^2V)/(dx^2))_{x=y/6}=-8y+24*(y/6)`

`=-4y<0`

S0 x=y/6 ,will give maximum volume of the box.

For validity please substitute the values of y and x separately your self.

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