Answer:
Let the side of the square to be cut-off be x cm\[(0<x<9)\]. Then, the length and the breadth of the box will be \[(18-2x)\]cm each and the height of the box is x cm Let V the volume of the open box formed by folding up the flaps, then \[V=x\,(18-2x)\,\,(18-2x)\] \[=4x\,{{(9-x)}^{2}}=4x\,(81+{{x}^{2}}-18x)\] \[=4\,({{x}^{3}}-18{{x}^{2}}+81x)\] On differentiating twice w.r.t. x, we get \[\frac{dV}{dx}=4\,(3{{x}^{2}}-36x+81)=12\,({{x}^{2}}-12x+27)\] And \[\frac{{{d}^{2}}V}{d{{x}^{2}}}=12\,(2x-12)=24\,(x-6)\] For maxima or minima, put\[\frac{dV}{dx}=0\] \[\Rightarrow \] \[12\,({{x}^{2}}-12x+27)=0\] \[\Rightarrow \] \[{{x}^{2}}-12x+27=0\Rightarrow (x-3)\,\,(x-9)=0\] \[\Rightarrow \] \[x=3,\,\,9\] But \[x=9\] is not possible. \[\therefore \] \[2x=2\times 9=18\] which is equal to side of square piece. At\[x=3,\,\,{{\left( \frac{{{d}^{2}}V}{d{{x}^{2}}} \right)}_{x\,\,=\,\,3}}=24\,(3-6)=-72<0\] \[\therefore \]By second derivative test, \[x=3\]is the point of maxima. Hence, if we cut-off the side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible. For maximum volume, Length and breadth of the box\[=18-2x\] \[=18-2\times 3=12\,\,cm\] and height of the box\[=x=3\,\,cm\] \[\therefore \]Maximum volume of the box \[=12\times 12\times 3=432\,\,c{{m}^{3}}\]
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