Question

a clothing store had a rectangular clearance section with a length that is twice the width w. during a sale, the secrion is expanded to an area of 2w^2 + 19w+35ft^2. find the amount of the increase in the length and width of the clearance section

Answer 1

The length is increased by 5 feet and width is increased by 7 feet.

Explanation:

Given:

Length of section is twice the width.

During sale the section is expanded to an area =
`(2w^2+19w+35)\ ft^2`

To find increase in length and width.

Solution:

Let the length of the section be =
`l\ ft`

width of the section =
`w\ ft`

Expression for area :

`2w^2+19w+35`

Factoring:

`2w^2+14w+5w+35`

`2w(w+7)+5(w+7)`

`(2w+5)(w+7)`

So, area of the section after increase can be given as
`(2w+5)(w+7)`

We know that length is twice the width, which means:

`l=2w`

Substituting the value of
`2w` in the factored expression of area.

`(l+5)(w+7)`

Since area of triangle is product of length and width, so we have:

New length of section =
`(l+5)\ ft`

New width of the section =
`(w+7)\ ft`

Thus, length is increased by 5 feet and width is increased by 7 feet.