Question

A square tin sheet of side 12 inches is converted into a box with open top in the following steps- The sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an interger, then what value of x maximizes the volume of the box?

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

When we fold the sheet in the form of a box it becomes a cuboid whose length is $12-2x$, width is $12-2x$ and height is $x$
So volume $\left(V\right)=l\times b\times h=(12-2x{)}^2\times x$
$\therefore V=144x-48x^2+4x^3$
For maximum volume of the box, $\frac{dV}{dx}=0$
$\therefore \frac{dV}{dx}=144-96x+12x^2=0$
$\therefore (x-2)(x-6)=0$
$\therefore x=2orx=6$
At $x=2$, $V=[12-2(2){]}^2\times 2=128$........(max)
At $x=6$, $V=[12-2(6){]}^2\times 6=0$