Question

A box is constructed from a rectangular metal sheet of $21$ cm. by $16$ cm, by cutting equal squares of sides $x$ from the corners of the sheet and then turning up the projected portions. For what values of $x$ the volume of the box will be maximum?

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is B 3

Let the dimensions of the box after cutting equal squares of side $x$ on the corner will be

$21-2x,16-2x$ and height $x$

$V=x(21-2x)(16-2x)$

$V=x(336-74x+4x^2)$

$V=4x^3-74x^2+336x$

$\frac{dV}{dx}=12x^2-148x+336$

For maximum or minimum,

$\frac{dV}{dx}=0$

$\frac{dV}{dx}=12x^2-148x+336=0$

or $3x^2-37x+84=0.$

$\therefore (x-3)(3x+28)=0$

$\therefore x=3$ ($x=-\frac{28}{3}$, not possible )

Now, $\frac{d^{2}V}{d{x}^2}=6x-37=18-37$

$=-19=-ive$ for $x=3.$

Hence V is maximum when $x=3$