Question

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting-off square from each corner and folding up the flaps. What should be the side of the square to be cut-off so that the volume of the box is maximum?

Answer 1

Let the side of the square to be cut off be x cm. Then, the height of the box is x, the length is 45-2x and the breadth is 24-2x.
Let V be the corresponding volume of the box then,

$V=x(24-2x)(45-2x)\Rightarrow V=x(4x^2-138x+1080)=4x^3-138x^2+1080x$
On differentiating twice w.r.t.x, we get
$\frac{dV}{dx}=12x^2-276x+1080$
and $\frac{d^{2}V}{d{x}^2}=24x-276$
For maxima put $\frac{dV}{dx}=0$
$\Rightarrow 12x^2-276x+1080=0\Rightarrow x^2-23x+90=0\Rightarrow (x-18)(x-5)=0\Rightarrow x=5,18$
It is not possible to cut-off a square of side 18 cm from each corner of the rectangular sheet. Thus, x cannot be equal to 18.
At $x=5,{\left(\frac{d^{2}V}{d{x}^2}\right)}_{x=5}=24\times 5-276=120-276=-156<0$
$\therefore$ By second derivative test, x=5 is the point of maxima.
Hence, the side of the square to be cut-off to make the volume of the box maximum possible is 5 cm.