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# A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.

$\begin{array}{1 1} 5\;cm \\ 4\;cm \\ 3\;cm \\ 6\;cm \end{array}$

Toolbox:
• Volume of a box=$l\times b\times h$
• $\large\frac{d}{dx}$$(xy)=x.\large\frac{d}{dx}$$(y)+y.\large\frac{d}{dx}$$(x) Step 1: Let each side of the square be cut off be x cm. For the box: Length=18-2x breadth=18-2x height=x Volume=x(18-2x)^2 On differentiating with respect to x we get \large\frac{dv}{dx}$$=x.2(18-2x)(-2)+(18-2x)^2.1$
$\quad=(18-2x)(-4x+18-20)$
$\quad=(18-2x)(18-6x)$
Step 2:
For maxima and minima
$\large\frac{dV}{dx}$$=0 \Rightarrow (18-2x)(18-6x)=0 \Rightarrow 324-108x-36x+12x^2=0 12x^2-144x+324=0 \Rightarrow 12[x^2-12x+27]=0 \Rightarrow [x^2-12x+27]=0 \Rightarrow x(x-9)-3(x-9)=0 \Rightarrow (x-9)(x-3)=0 x=3,9 Step 3: But x=9cm is not possible. On differentiating with respect to x we get Also \large\frac{d^2V}{dx^2}$$=(18-2x)(-6)+(18-6x)(-2)$
At $x=3$
$\large\frac{d^2V}{dx^2}$$=(18-6)(-6)+(18-18)(-2)$
$\qquad=12(-6)+0(-2)$
$\qquad=-72 <0$
Hence volume is maximum when $x=3$ (i.e) Square of side=3 cm is cut from each corner.
edited Aug 30, 2013