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Question

Mathematics Question on Application of derivatives

A square piece of tin of side 24cm24\, cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum?

A

22

B

44

C

66

D

88

Answer

44

Explanation

Solution

Let xx cm be the length of a side of the square which is cut-off from each corner of the plate. Then, sides of the box as shown in figure are 242x,242x24 - 2x, 24 - 2x and xx. Let V V be the volume of the box. Then, V=(242x)2x=4x396x2+576xV = (24 - 2x) ^2 x = 4\,x ^3 - 96\,x ^2 + 576\,x dVdx=12x2192x+576\Rightarrow \frac{dV}{dx} = 12x^{2} -192x + 576 and d2Vdx2\frac{d^{2}V}{dx^{2}} =24x192 = 24x - 192 For maximum or minimum values of VV, we must have dVdx=0\frac{dV}{dx} = 0 12x2192x+576=0 \Rightarrow 12 x^{2} - 192 x + 576 = 0 x216x+48=0 \Rightarrow x^{2} -16 x + 48 =0 (x12)(x4)=0 \Rightarrow \left(x-12\right)\left(x-4\right) = 0 x=12,4 \Rightarrow x= 12, 4 But, x=12x= 12 is not possible Therefore, x=4x=4. and [d2Vdx2]x=4=96<0\left[ \frac{d^{2}V}{dx^{2}}\right]_{x=4} = -96 <0 Hence, volume is maximum when x=4x = 4.