**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 the maximum possible?

**Solution:**

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 − 2*x,* and the breadth is 24 − 2*x*.

Therefore, the volume *V*(*x*) of the box is given by,

$\begin{aligned} V(x) &=x(45-2 x)(24-2 x) \\ &=x\left(1080-90 x-48 x+4 x^{2}\right) \\ &=4 x^{3}-138 x^{2}+1080 x \\ \therefore V^{\prime}(x) &=12 x^{2}-276 x+1080 \\ &=12\left(x^{2}-23 x+90\right) \\ &=12(x-18)(x-5) \\ V^{\prime \prime}(x) &=24 x-276=12(2 x-23) \end{aligned}$

Now, $V^{\prime}(x)=0 \Rightarrow x=18$ and $x=5$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.

∴*x* = 5

Now, $V^{\prime \prime}(5)=12(10-23)=12(-13)=-156<0$

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.