Advertisements
Advertisements
प्रश्न
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, in cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?
Advertisements
उत्तर
\[\text { Suppose square of side measuring x cm is cut off } . \]
\[\text { Then, the length, breadth and height of the box will be } \left( 45 - x \right), \left( 24 - 2x \right) \text { and } x, respectively . \]
\[ \Rightarrow \text { Volume of the box, V } = \left( 45 - 2x \right)\left( 24 - 2x \right)x\]
\[ \Rightarrow \frac{dV}{dx} = \left( 45 - 2x \right)\left( 24 - 2x \right) - 2x\left( 45 - 2x \right) - 2x\left( 24 - 2x \right)\]
\[\text { For maximum or minimum values of V, we must have }\]
\[\frac{dV}{dx} = 0\]
\[ \Rightarrow \left( 45 - 2x \right)\left( 24 - 2x \right) - 2x\left( 45 - 2x \right) - 2x\left( 24 - 2x \right) = 0\]
\[ \Rightarrow 4 x^2 + 1080 - 138x - 48x + 4 x^2 + 4 x^2 - 90x = 0\]
\[ \Rightarrow 12 x^2 - 276x + 1080 = 0\]
\[ \Rightarrow x^2 - 23x + 90 = 0\]
\[ \Rightarrow x^2 - 18x - 5x + 90 = 0\]
\[ \Rightarrow x\left( x - 18 \right) - 5\left( x - 18 \right) = 0\]
\[ \Rightarrow x - 18 = 0 \text{ or }x - 5 = 0\]
\[ \Rightarrow x = 18 \text{ or }x = 5\]
\[\text { Now,} \]
\[\frac{d^2 V}{d x^2} = 24x - 276\]
\[ \frac{d^2 V}{d x^2}_{x = 5} = 120 - 276 = - 156 < 0 \]
\[ \frac{d^2 V}{d x^2}_{x = 18} = 432 - 276 = 156 > 0\]
\[\text { Thus, volume of the box is maximum when x } = 5 cm . \]
\[\text { Hence, the side of the square to be cut off measures } 5 cm .\]
