Question

An open topped box is to be constructed by removing equal squares from each corner of a $3 m$ by $8 m$ rectangular sheet of aluminium and folding up the sides. The largest volume of such a box is

• A. $\frac{200}{27} m^3$
• B. $\frac{100}{27} m^3$
• C. $\frac{90}{31} m^3$
• D. $\frac{81}{28} m^3$

Answer 1

Answer: (A)

Let $x m$ be the length of a side of the removed squares. Then, the height of the box is $x$, length is $8-2 x$ and breadth is $3-2 x$. If $V(x)$ is the volume of the box, then

Therefore, $V(x) =x(3-2 x)(8-2 x)$
$ =4 x^3-22 x^2+24 x $
$\Therefore V^{\prime}(x) =12 x^2-44 x+24=4(x-3)(3 x-2) $
$ V^{\prime}(x) =24 x-44$
Now, $V^{\prime}(x)=0$ gives $x=3, \frac{2}{3}$. But $x \neq 3$
Thus, we have $x=\frac{2}{3}$. Now $V^{\prime \prime}\left(\frac{2}{3}\right)=24\left(\frac{2}{3}\right)-44=-28 < 0$.
Therefore, $x=\frac{2}{3}$ is the point of maxima, i.e., if we remove a square of side $\frac{2}{3} m$ from each corner of the sheet and make a box from the remaining sheet, then the volume of the box such obtained will be the largest and it is given by
$V\left(\frac{2}{3}\right)=4\left(\frac{2}{3}\right)^3-22\left(\frac{2}{3}\right)^2+24\left(\frac{2}{3}\right)=\frac{200}{27} m ^3$