a rectangle is 3 times as wide as a square. the rectangle's length is 15 ft shorter than 5 times the square's length. Given that the sum of the two perimeters must be less than 150 ft, find the set of all possible lengths for the rectangle

I see from the first statement that since the rectangle length is 15ft less than five times the sqare side, the square side cannot be less than 3.

lets make it 3.01 for the square. Now, the width of the rectangle then will be 3.01x3, or 9.03 width, and length .05.

So that is the smallest rectangle and the smallest square.

Now the largest: Perimerter sum <150

150>4s+2w+2l
150>4s+2*3s + 2*(15-5s)
solve for s, then w, then l.

To find the set of all possible lengths for the rectangle, we need to understand the relationship between the rectangle and the square. Let's start by assigning variables to the unknown values.

Let:
- Width of the square be x
- Length of the square be also x
- Width of the rectangle be 3x (since it is 3 times wider than the square)
- Length of the rectangle be 5x - 15 (15 ft shorter than 5 times the square's length)

Now, let's calculate the perimeters of the square and the rectangle.

Perimeter of a square = 4 * side length
Perimeter of a rectangle = 2 * (length + width)

For the square:
Perimeter of the square = 4x

For the rectangle:
Perimeter of the rectangle = 2 * (5x - 15 + 3x)
= 2 * (8x - 15)
= 16x - 30

According to the problem, the sum of the two perimeters must be less than 150 ft.

So, we have the inequality:

4x + (16x - 30) < 150

Simplifying the inequality, we get:

20x - 30 < 150
20x < 180
x < 9

Therefore, the width of the square, x, must be less than 9.

Since the width of the rectangle is 3 times the width of the square, the width of the rectangle, 3x, must also be less than 9.

So, the set of all possible lengths for the rectangle is any value less than 9 times 3, which is 27 ft.

In conclusion, the set of all possible lengths for the rectangle is any value less than 27 ft.