# a rectangular garden has a flowerbed in the middle. Between the boundary of the garden and the flowerbed is a path, the same width all the way round. The area of the flowerbed is half that of the...

a rectangular garden has a flowerbed in the middle. Between the boundary of the garden and the flowerbed is a path, the same width all the way round. The area of the flowerbed is half that of the garden. What are the dimensions of the flowerbed and the garden, if all the lengths (including the width of the path) are intergers?

ishpiro | College Teacher | (Level 1) Educator

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Suppose the flowerbed has length a and width b. Since the path around the flowerbed has the same width all the way around (let's call it w), the garden has dimensions a + 2w and b + 2w.

The area of the garden is (a +2w)(b + 2w) and is twice the area of the flowerbed, ab:

`(a+2w)(b+2w) = 2ab`

`ab+2bw+2aw +4w^2 = 2ab`

`4w^2 + 2(a+b)w - ab=0`

`w^2 + (a+b)/2 w - (ab)/4 = 0`

This is a quadratic equation. To produce integer values of w, ab has to be a multiple of 4 and a and b have to be such that the quadratic trinomial on the left side have to be factorable. (Which means there have to be a pair of factors of ab/4 such that their DIFFERENCE equals to (a+b)/2.)

Let's go through some multiples of 4:

ab = 4; then ab/4 = 1; not possible

ab = 8; then ab/4 = 2; not possible

ab = 12; then ab/4 = 3, (a + b)/2 would have to be 2. No pair of factors of 12 will result in half-sum of 2.

ab = 16; then ab/4 = 4, (a + b)/2 would have to be 3. No pair of factors of 16 will result in half-sum of 2.

ab = 20; then ab/4 = 5, (a + b)/2 would have to be 4. No pair of factors of 20 will result in half-sum of 4.

ab = 24; then ab/4 = 6, (a + b)/2 would have to be 5 OR 1.

a = 4 and b = 6: (4+6)/2 = 5 !!!

The equation becomes

`w^2 + 5w - 6 = 0`

`(w+6)(w-1) = 0`

Only positive solution makes sense in the context of the problem:

w = 1

Dimensions of the flowerbed is 4 by 6, and the dimensions of the garden are 6 by 8.

Check: 24 = half of 48 - it works!