You have just planted a rectangular flower bed of red roses in a city park. You want to plant a border of yellow roses around the flower bed. Because you bought the same number of yellow roses as red roses, the area of the flower bed and border will be equal.

What should the width of the border of the yellow roses be?

Edit: Picture showed original flower bed to be 8ft by 12ft
Last edited by: J-No: Sep 12, 17 19:11
Just to be clear, is the border going around the original 8x12 flower bed and you are looking for two areas to be equal, that is the area of the border and the area of the original bed?
Last edited by: DJRed: Sep 12, 17 19:31
The way that you explained it, you have 8x12 => 96 sq. ft. of red roses. The total area will be twice that, 192 sq. ft. Which means that the new plot would be 12x16, with a two foot border around the original plot.

Of course, you are going to need to use variables in 10th grade.

Let x be the width of the yellow border (it isn't stated, but the border needs to be equal all around, or there are infinite solutions). Drawing it, you will see 8 basic segments, 2 widths, 2 lengths, 4 corners (you can divvy up the segments differently). The area of the yellow bed is the result of all of those areas added together:

2w*x+2l*x + 4x^2 = 96

40x + 4x^2 = 96

0 = x^2 + 10x - 24
0 = (x+12)(x- 2)

checking each factor, and solving for zero, {x = -12} isn't a real answer, so x = 2.
Last edited by: oldandslow: Sep 12, 17 19:47
oldandslow wrote:
The way tha tyou explained it, you have 8x12 => 96 sq. ft. of red roses. The total area will be twice that, 192 sq. ft. Which means that the new plot would be 12x16, with a two foot border around the original plot.

^This is the way I read it.

You'll need to solve a quadratic equation. Let x = the width of the border. If the inner border is 8x12 it has an area of 96, but that also means the entire area of the rectangle created by the new border is (x + 12 + x) times (x + 8 + x). But this is the entire new area, not just the area of the new border (which is also 96). That means the area of the new border is (x + 12 + x) times (x + 8 + x) minus 96. I hope this makes sense because a picture would make this so much easier.

Then you just solve (x + 12 + x)(x + 8 + x) - 96 = 96. In words, the area of the new rectangle created less the area of the already existing inside rectangle equals the area of the new border.

(2x+12)(2x+8)-96=96
(2x+12)(2x+8)=192
4x^2+40x+96=192
4x^2+40x-96=0
x^2+10x-24=0
(x+12)(x-2)=0
x= -12 or x = 2

So, x =2 and the width of the border is 2.
Thanks all!!!

I'm an idiot.