Question

You are constructing a box from a piece of cardboard with dimensions 3 by 5 meters. You cut equal-size squares from each corner of the cardboard so you may fold the edges to construct the open top box. What are the dimensions of the box with the largest volume?

Answer 1

The dimension of the box are 2.878 m by 4.878 m by 0.061 m.

Explanation:

Given that,

A box is constructed from a piece of cardboard with dimensions 3 m by 5 m.

Assume the length of side of the square be x.

Then the length of the box= (3-2x) m

The width of the box is =(5-2x) meter.

The height of the box is = x m.

Then the volume of the box =length×width×height

=(3-2x)(5-2x)x cubic meters

=15x-16x²+4x³

Let,

V=15x-16x²-4x³

Differentiating with respect to x

V'= 15 - 32x+12x²

Again differentiating with respect to x

V''= -32+24x

At first we have to find out at which point the volume of the box maximum.

For that, we set V'=0

15 - 32x+12x²=0

Applying quadratic formula
`x=(-b\pm√(b^2-4ac))/(2a)` , here a= 12, b= -32 and c=15

`\therefore x=(-(-32)\pm√((-32)^2-4.12.15))/(2.12)`

=2.06,0.61

So, x= 2.06 does not possible, because the length (3-2x) becomes negative.

∴x= 0.061 m

Since
`V'|_(x=0.061)'=-32+24(0.061)<0`, the volume of the maximum at x =0.061 m.

The length of the box is= {3- 2(0.061)} m= 2.878 m

The width of the box is ={5-2(0.061) }meter= 4.878 m

The height of the box is = 0.061 m.

The dimension of the box are 2.878 m by 4.878 m by 0.061 m.