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Question

# A square tin sheet of side 12 inches is converted into a box with open top in the following steps- The sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an interger, then what value of x maximizes the volume of the box?

A
3
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B
4
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C
1
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D
2
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Solution

## The correct option is C 2When we fold the sheet in the form of a box it becomes a cuboid whose length is 12−2x, width is 12−2x and height is xSo volume (V)=l×b×h=(12−2x)2×x∴V=144x−48x2+4x3For maximum volume of the box, dVdx=0∴dVdx=144−96x+12x2=0∴(x−2)(x−6)=0∴x=2orx=6At x=2, V=[12−2(2)]2×2=128........(max)At x=6, V=[12−2(6)]2×6=0

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