Answer:
Let the side of the square to be cut-off be x cm. Then, the length and breadth of the box will be \[(24-2x)\] cm each and the height of the box is x cm. Let V be the volume of the open box formed by folding up the flaps, then \[V=x(24-2x)(24-2x)\] \[=4x(12-{{x}^{2}})=4x(144+{{x}^{2}}-24x)\] \[=4({{x}^{3}}-24{{x}^{2}}+144x)\] On differentiating both sides twice w.r.t. x, we get \[\frac{dV}{dx}=4(3{{x}^{2}}-48x+144)\] \[\frac{dV}{dx}=12({{x}^{2}}-16x+48)\] and \[\frac{{{d}^{2}}V}{d{{x}^{2}}}=12(2x-16)=24(x-8)\] Now, for maximum or minimum, put \[\frac{dV}{dx}=0\] \[\Rightarrow \] \[12({{x}^{2}}-16x+48)=0\] \[\Rightarrow \] \[{{x}^{2}}-12x-4x+48=0\] \[\Rightarrow \] \[x(x-12)-4(x-12)=0\] \[\Rightarrow \] \[(x-12)(x-4)=0\] \[\Rightarrow \] x = 12 or x =4 But x = 12 is not possible, as length and breadth of box will become zero \[\therefore \] x = 4 Also, for x = 4, \[\frac{{{d}^{2}}V}{d{{x}^{2}}}=24(-\,4)<0\] \[\therefore \] By second derivative test, x = 4 is point of maxima. Hence, if we cut-off the side 4 cm from each corner the square place and make a box from the remaining piece, then the volume of the box obtained is the largest possible. Value As our country is still developing and most of the Indian people are from the middle class, so we should utilise our resources in proper way.
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