A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top,

Solution:

Suppose square of side measuring $x \mathrm{~cm}$ is cut off.

Then, the length, breadth and height of the box will be $(45-x),(24-2 x)$ and $x$, respectively.

$\Rightarrow$ Volume of the box, $V=(45-2 x)(24-2 x) \mathrm{x}$

$\Rightarrow \frac{d V}{d x}=(45-2 x)(24-2 x)-2 x(45-2 x)-2 x(24-2 x)$

For maximum or minimum values of $\mathrm{V}$, we must have

$\frac{d V}{d x}=0$c

$\Rightarrow(45-2 x)(24-2 x)-2 x(45-2 x)-2 x(24-2 x)=0$

$\Rightarrow 4 x^{2}+1080-138 x-48 x+4 x^{2}+4 x^{2}-90 x=0$

$\Rightarrow 12 x^{2}-276 x+1080=0$

$\Rightarrow x^{2}-23 x+90=0$

$\Rightarrow x^{2}-18 x-5 x+90=0$

$\Rightarrow x(x-18)-5(x-18)=0$

$\Rightarrow x-18=0$ or $x-5=0$

$\Rightarrow x=18$ or $x=5$

Now,

$\frac{d^{2} V}{d x^{2}}=24 x-276$

$\left.\frac{d^{2} V}{d x^{2}}\right|_{x=5}=120-276=-156<0$

$\left.\frac{d^{2} V}{d x^{2}}\right|_{x=18}=432-276=156>0$

Thus, volume of the box is maximum when $x=5 \mathrm{~cm}$.

Hence, the side of the square to be cut off measures $5 \mathrm{~cm}$.