Question

A rectangular sheet of tin $45cm$ by $24cm$ 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 $xcm$ length is cut from one corner,
Length after cutting
$=45-x-x=45-2x$
Width after cutting
$=24-x-x=24-2x$
Height of the box $=x$
Volume of the box is
$V=(45-2x)(24-2x)x$
$\Rightarrow V=2x(2x-45)(x-12)$
$\Rightarrow V=2x(2x^2-69x+540)$
$\Rightarrow V=2x(2x^3-69x^2+540x)$
Differntiating w.r.t $x$
$\frac{dV}{dx}=2(6x^2-138x+540)$
$\Rightarrow \frac{dV}{dx}=12(x^2-23x+90)$
$\Rightarrow \frac{dV}{dx}=12(x-5)(x-18)$
Putting $\frac{dV}{dx}=0$
$12(x-5)(x-18)=0$
$\Rightarrow x=5,18$
These are the critical points.
Again, Differentiating w.r.t $x$
$\frac{d^{2}V}{d{x}^2}=12(2x-23)$
At $x=5$, we get
$\frac{d^{2}V}{d{x}^2}=12(-13)<0$
At $x=18$, we get
$\frac{d^{2}V}{d{x}^2}=12\left(13\right)>0$

For maxima, $\frac{d^{2}V}{d{x}^2}<0$

Hence, $5cm$ should be cut off from the side of the square so that the volume of the box is maximum.