A square piece of tin of side 18 cm is to be made

Solution:

Let the side of the square to be cut off be x cm.
Then, the length and the breadth of the box will be (18 − 2x) cm each and height of the box will be x cm.

Volume of the box, $V(x)=x(18-2 x)^{2}$

$V^{\prime}(x)=(18-2 x)^{2}-4 x(18-2 x)$

$=(18-2 x)(18-2 x-4 x)$

$=(18-2 x)(18-6 x)$

 

$=12(9-x)(3-x)$

$V^{\prime \prime}(x)=12(-(9-x)-(3-x))$

$=-12(9-x+3-x)$

 

$=-24(6-x)$

For maximum and minimum values of $V$, we must have

$V^{\prime}(x)=0$

$\Rightarrow x=9$ or $x=3$

If $x=9$, then length and breadth will become 0

$\therefore x \neq 9$

$\Rightarrow x=3$

Now,

$V^{\prime \prime}(3)=-24(6-3)=-72<0$

$\therefore x=3$ is the point of maxima.

$V(x)=3(18-6)^{2}=3 \times 144=432 \mathrm{~cm}^{3}$

Hence, if we remove a square of side $3 \mathrm{~cm}$ from each corner of the square tin and make a box from the remaining sheet, then the volume of the box so obtained would be the largest, i.e. $432 \mathrm{~cm}^{3}$.