Question

Suppose you want to make an open box out of a piece of card board by cutting small squares at the four corners and folding up the sides. If the piece of card board is a square whose sides are 1 m. long, how big a square should you cut from the corners to maximize the volume of the box?

Answer 1

Answer:

The box of length $x=\frac{1}{6}$ must be cut from each corner to maximize the volume.

Step-by-step explanation:

It is given that the side of a square card board is 1 m.

Let small squares of length x are cut from the four corners. The dimensions of the box are

Length = (1-2x) m

Width = (1-2x) m

Height = x m

The volume of a cuboid is

$V=l\times b\times h$

where, l is length and b is breadth and h is height.

The volume of a rectangular box.

$V=(1-2x)(1-2x)x$

$V=(1-2x)(x-2x^2)$

$V=x-2x^2-2x(x-2x^2)$

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

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

We need to maximize the volume. Differential above equation with respect to x.

$V'=12x^2-8x+1$                .... (1)

Equate V'=0 to find the critical points.

$12x^2-8x+1=0$

$12x^2-6x-2x+1=0$

$6x(2x-1)-1(2x-1)=0$

$(6x-1)(2x-1)=0$

$x=\frac{1}{6},\frac{1}{2}$

Differential equation (1) with respect to x.

$V''=24x-8$

At $x=\frac{1}{6}$

$V''=24(\frac{1}{6})-8=-4<0$

Volume is maximum at $x=\frac{1}{6}$.

At $x=\frac{1}{2}$

$V''=24(\frac{1}{2})-8=4>0$

Volume is minimum at $x=\frac{1}{6}$.

Therefore the box of length $x=\frac{1}{6}$ must be cut from each corner to maximize the volume.