question_answer

An expensive square piece of golden colour board of side 24 cm is to be made into a box of without top by cutting a square from each corner and folding the flaps to form a box. What should be the side of the square piece to be cut from each corner of the board to hold maximum volume and minimise the wastage? What is the importance of minimising the wastage in utilising the resources?

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.