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i have a garden that measures 40x50 feet. i want to add a tile walkway around it. how wide can it be if i have 784 square feet of stones?

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2 Answers

The perimeter of the garden is 180 feet. The length is 50 and the width is 40. You have 784 square foot tiles for a walkway around the perimeter.

You will need 184 tiles to go around the perimeter one time, because you must extend the length by one tile on each end of the lengths so that all the tiles for will butt up against each other (52+52+40+40=184). You now have a walkway that is one tile deep, with 600 tiles left.

Since you have 600 tiles left over, you can keep adding to the width of the walkway. Let's add another layer of tiles all the way around, using up 192 tiles. We had to add another tile to each end of the new length (52+2=54) plus a tile at each end of the length “going up” to get us up and over to the next row (54+2=56) for a total of 192 tiles (56+56+40+40=192). We now have a double tiled walkway, with 408 tiles left over.

Adding a third layer of 200 tiles leaves us with 208 tiles. We had to add a tile at each end of the new length (56+2=58), plus a tile at each end of the length “going up” to get us up and over to the next row (58+2=60) for a total of 200 tiles (60+60+40+40=200).

We can add a fourth layer of tiles in the same manner (64+64+40+40=208), using up our remaining 208 tiles, with none left over.

So, we were able to build a 4 foot walkway around our garden using all of the 784 tiles.

Once you're done building the walkway, you'll have two rectangles, one (the garden) inside the other (the outer edge of the walkway).

Recall that to find the area of a rectangle we use the formula A = bh, where A is the area, b is the length of the base of the rectangle, and h is the height of the rectangle.

The area of the garden is given by

40ft² × 50ft² = 2000ft²

Let w be the width of the walkway.  Since the walkway goes along each edge of the garden, its width will be added twice to each dimension of the garden.  The outer rectangle will have a base of 2w + 40ft and a height of 2w + 50ft.

The widest walkway we can create will have an outer rectangle whose area is the sum of 784ft² (the tiles available) and 2000ft² (the area we found for the inner rectangle).  Using all of the information we've accumulated, we can find the maximum width of the rectangle by using the inequality

(2w + 40ft) × (2w + 50ft) ≤ 2,784ft²

We can simplify the left-hand side using the FOIL method.

(2w × 2w) + (2w × 40) + (2w × 50) + (40 × 50) ≤ 2,784

4w² + 80w + 100w + 2,000 ≤ 2,784

4w² + 180w + 2,000 ≤ 2,784

To solve that last inequality, we first zero out the right-hand side, by subtracting 2,784 from both sides. 

4w² + 180w + -784 ≤ 0

I'll leave it to you to use the quadratic formula on that last equation, but its roots are at 4 and -49.  Only the positive root makes sense in our problem (because the walkway can't have negative width), so the widest our walkway can be is 4ft.