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

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- 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.

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