Robin is mowing a rectangular field that measures 24 yards by 32 yards, by pushing the mower around and around the outside of the plot. This creates a widening border that surrounds the un-mowed grass in the center. During a brief rest, Robin wonders whether the job is half done yet. How wide is the uniform mowed border when Robin is half done?

We also are going to have this patch of grass be directly in the center of the original field. So from the center there will be the same amount of space from the top to the top edge as the bottom to the bottom edge and the sides as well. Like a picture frame almost.

So this distance from the edge to the patch let's call x. This also mean thatthe starting 24 feet is going to take x off of both ends, so that means we'll have 24-2x and then 32-2x as the new dimensions. Let me know if this doesn't make sense. x is basically how wide the border is, or how wide the "frame" is, using that comparison.

Anyway, we have the new dimensions and area of the patch so let's do some algebra.

(24-2x)(32-2x)=384

768 - 48x - 64x + 4x^2 - 384 = 0

4x^2 - 112x + 384 = 0

Now we have it in the form of a standard quadratic. There are numerous ways to solve for x, but I am going to complete the square. Let me know if you don't understand a method to solve for x

4x^2 - 112x + 384 = 0

4(x^2 - 28x + 96) = 0

4((x - 14)^2 -100) = 0

4(x-14)^2 - 400 = 0

4(x-14)^2 = 400

(x-14)^2 = 100

x - 14 = +/- 10 (can be positive or negative)

x = +/- 10 + 14

x = 4, 24

Since we have two numbers we have to check both. You may see a problem right away with the larger one. if we took 24 away from 24, that means there would be no grass left in between, so it can't be that one. Which leaves us with x=4

This means the new dimensions are 24-2*4 = 16 and 32-2*4=24

Then checking 16*24 = 384, so we got it.

So the answer about how wide is 4, the border/ frame is 4 yards wide.

Let me know if this didn't make sense or some part was hard to understand.