question_answer

Square piece of tin of side 18 cm is to be made into a box without a 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 be maximum? Also, find the maximum volume of the box.

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                                     \[=12\times 12\times 3=432\,\,c{{m}^{3}}\]