From a rectangular sheet of dimensions $30cm \times 80cm$ four equal squares of side $x cm$. are removed at the corners, and the sides are then turned up so as to form an open rectangular box. Find the value of $x$, so that the volume of the box is the greatest.

A rectangular sheet of dimensions $30cm \times 80cm$ Given, A square of side ‘$x$’ cm are removed at corners and turned up to form a box

Now length of box $\left(l\right)=80-2x$

Breadth of box $\left(b\right)=30-2x$

Height of box $\left(h\right)=x$

The volume of box $\left(v\right)=lbh$

$=(80-2x)(30-2x)x$

$=\left(2400-160x-60x+4x^2\right)x$

$=2400x-220x^2+4x^3$

Let $f\left(x\right)=4x^3-220x^2+2400x$

$f^{\mathrm{\prime }}\left(x\right)=12x^2-440x+2400;f^{\prime \prime }\left(x\right)=24x-440$

Consider $f^{\mathrm{\prime }}\left(x\right)=0$

$12x^2-440x+2400=0\Rightarrow 3x^2-110x+600=0$

$3x^2-90x-20x+600=0$

$3x(x-30)-20(x-30)=0$

$(x-30)(3x-20)=0\Rightarrow x=30,20/3$

$x=30$ is not possible

$\left[\begin{array}{l}\because b=30-2x\\ =30-2\left(30\right)=30-60=-30\end{array}\right]; x=\frac{20}{3}$

$f^{\prime \prime }\left(x\right)=24\left(\frac{20}{3}\right)-440=160-440=-280<0$

$f\left(x\right)$ is (maximum)

Volume of the box is maximum when $x=\frac{20}{3}\mathrm{cm}$