Question

A manufacturing company is building a rectangular room in their warehouse to store their products The length of the room is 1 more than 3 times its width. The area of the room is 80 square meters What are the dimensions of the room?

Answer 1

See a solution process below:

Explanation:

First, let's call the length of the room: `l`

And, let's call the width of the room `w`

Then. we can write the relationship of the length of the room to the width of the room as:

`l = 3w + 1`

The formula for the Area of a rectangle is:

`A = l xx w`

We can substitute:

• `80` for `A`
• `(3w + 1)` for `l`

and solve for `w`

`A = l xx w` becomes:

`80 = (3w + 1) xx w`

`80 = (3w + 1)w`

`80 = 3w^2 + w`

`80 - color(red)(80) = 3w^2 + w - color(red)(80)`

`0 = 3w^2 + w - 80`

`0 = (3w + 16)(w - 5)`

Now, solve `(w - 5)` for `0`

`w - 5 = 0`

`w - 5 + color(red)(5) = 0 + color(red)(5)`

`w - 0 = 5`

`w = 5`

We do not need to solve the other term for `0` because it will be a negative result and you cannot have a negative width for a room.

To find the length substitute `5` for `w` in the equation for the relationship for the length and the width and calculate `l`:

`l = 3w + 1` becomes:

`l = (3 xx 5) + 1`

`l = 15 + 1`

`l = 16`

The room is `16` meters by `5` meters

Answer 2

Analize the relationships between the length and the width.

Explanation:

The area of a rectangle is width times length.

But let’s re-word the length in terms of the width. The length is one more than the triple of the width.

So the area of this particular rectangle is:

width · ( 3 · width + 1)

And, since the area is given, we can say that:

width · ( 3 · width + 1) = 80 square meters

Applying the distributive property:

3 · `width^2` + width = 80

Let’s prepare this equation for factoring by subtracting 80 to both sides:

3 · `width^2` + width - 80 = 0

One way of factoring the left side is to find two numbers (m and n) that give a product of (3)(-80)=-240 and a total of 1:

m · n = -240

m + n = 1

Since the product is negative, one of the two unknown numbers must be positive and the other one must be negative.

Also, since their addition is 1, the positive number must be just one unit greater than the negative number.

Let’s try with -10 and 11.

-10 · 11 = -110

We need bigger numbers.

How about -14 and 15?

-14 · 15 = -210

We need something bigger than that.

Say, -16 and 17.

-16 · 17 = -272

Oops! We went too far.

So, -15 and 16?

-15 · 16 = -240

There!

Now, we can express

3 · `width^2` + width - 80 = 0

as

3 · `width^2` - 15 · width + 16 · width - 80 = 0

Let’s find a common factor for the first two terms and a common factor for the last two terms, and use them to factor the left side of the equation:

(3 · width) (width - 5) + 16(width - 5) = 0

Since (width - 5) is a common factor, let’s factor further:

(width - 5) (3 · width + 16) = 0

Two factors give us a product of zero. This can only be true if the first factor is zero or if the second factor is zero.

Let’s explore making the first factor a zero.

width - 5 = 0

`width_1` = 5

Let’s explore making the second factor a zero.

3 · width + 16 = 0

3 · width = -16

`width_2` = -16/3 = -5 and 1/3

This is theoretically correct, but makes no sense in a real-world warehouse.

Let’s confirm a width of 5 meters.

The triple of 5 is 15, so the length must be 16.

The area would be 5 · 16 = 80.

It works!