Question

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both sides and two-inch margins at the bottom and top, what are the dimensions of the smallest piece of paper (by area) that can be used? Choose 1 answer:

Answer 1

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Explanation:

We have that:

`Area = 128`

Let the dimension of the paper be x and y;

Such that:

`Length = x`

`Width = y`

So:

`Area = x * y`

Substitute 128 for Area

`128 = x * y`

Make x the subject

`x = (128)/(y)`

When 1 inch margin is at top and bottom

The length becomes:

`Length = x + 1 + 1`

`Length = x + 2`

When 2 inch margin is at both sides

The width becomes:

`Width = y + 2 + 2`

`Width = y + 4`

The New Area (A) is then calculated as:

`A = (x + 2) * (y + 4)`

Substitute
`(128)/(y)` for x

`A = ((128)/(y) + 2) * (y + 4)`

Open Brackets

`A = 128 + (512)/(y) + 2y + 8`

Collect Like Terms

`A = (512)/(y) + 2y + 8+128`

`A = (512)/(y) + 2y + 136`

`A= 512y^(-1) + 2y + 136`

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

`A' = -512y^(-2) + 2`

Set

`A' = 0`

This gives:

`0 = -512y^(-2) + 2`

Collect Like Terms

`512y^(-2) = 2`

Multiply through by
`y^2`

`y^2 * 512y^(-2) = 2 * y^2`

`512 = 2y^2`

Divide through by 2

`256=y^2`

Take square roots of both sides

`\sqrt{256=y^2`

`16=y`

`y = 16`

Recall that:

`x = (128)/(y)`

`x = (128)/(16)`

`x = 8`

Recall that the new dimensions are:

`Length = x + 2`

`Width = y + 4`

So:

`Length = 8 + 2`

`Length = 10`

`Width = 16 + 4`

`Width = 20`

To double-check;

Differentiate A'

`A' = -512y^(-2) + 2`

`A`

`A`

`A`

The above value is:

`A`

This means that the calculated values are at minimum.

Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches