The Garden Border Problem

The following problem is in the McDougal Littell Algebra 2 textbook and is a typical word problem for a section on solving quadratic equations: 

You have just planted a rectangular flower bed of red roses in a park near your home.  You want to plant a border of yellow roses around the flower bed as shown.  Since you bought the same number of red and yellow roses, the areas of the border and inner flower bed will be equal.  What should the width x of the border be? 

To solve this problem, you must first write an area equation.  The length of the whole garden is 12 feet plus the unknown widths of the left and right borders, which can be expressed as 2x + 12.  The width of the whole garden is 8 feet plus the unknown widths of the top and bottom borders, which can be expressed as 2x + 8.  The area of the whole garden is the area of the red rose garden (which is 8 feet times 12 feet or 96 feet squared) plus the area of the yellow rose garden (which is the same as the red rose garden or also 96 feet squared), which added together is 192 feet squared.  Since the area is length times width, the equation to solve is (2x + 12)(2x + 8) = 192.

The next step is to solve this area equation.  Multiplying (2x + 12)(2x + 8) gives us 4x² + 40x + 96, so 4x² + 40x + 96 = 192, and subtracting 192 to the left side gives us 4x² + 40x – 96 = 0, and dividing everything by the common factor 4 gives us x² + 10x – 24 = 0.  At this point, there are several methods for solving this quadratic (such as factoring, completing the square, quadratic equation, and graphing) but we will use the quadratic equation x = ^{-b ± √(b^2 – 4ac)} / _2a, where a = 1, b = 10, and c = -24.  Therefore, x = ^{-10 ± √(10^2 – 4·1·-24)} / _2·1 = ^{-10 ± √(100 + 96)} / ₂ = ^{-10 ± √196} / ₂ = ^{-10 ± 14} / ₂, which means x = -12 or x = 2.  Since x represents a geometrical dimension, it cannot be negative, and therefore the border width x must be 2 feet long.

You will notice that this answer conveniently comes out as an integer, and not as a decimal.  But what would happen if the problem started out with different dimensions for the inner garden?  Would the border width still be an integer?  Let us examine the same problem but with a starting inner garden of 8 feet by 8 feet, as pictured below:

  

This time both the length and the width of the whole garden can be expressed as 2x + 8, and the area of the whole garden is 2 times 8 feet by 8 feet, or 128 feet squared, giving us the equation (2x + 8)(2x + 8) = 128.  Multiplying (2x + 8)(2x + 8) gives us 4x² + 32x + 64 = 128, subtracting 128 to the left side gives us 4x² + 32x – 64 = 0, and dividing everything by the common factor 4 gives us x² + 8x – 16 = 0.  Using the quadratic equation x = ^{-b ± √(b^2 – 4ac)} / _2a, where a = 1, b = 8, and c = -16 gives us x = ^{-8 ± √(8^2 – 4·1·-16)} / _2·1 = ^{-8 ± √(64 + 64)} / ₂ = ^{-8 ± √128} / ₂ = ^{-8 ± 8√2} / ₂ = -4 ± 4√2.  Since x cannot be negative, the border width must be -4 + 4√2 feet long, which is not an integer answer.

Can we come up with different dimensions for the inner garden such that the border width solution comes out as an integer?  We already know one solution set is (8, 12, 2) from the original problem, and using the properties of proportions and dividing each number by two we can also include (4, 6, 1).  In fact, using the same argument we can include all solution sets in the form of (4k, 6k, k) where k is any positive integer.  To simplify things, we will say that (4, 6, 1) is a “garden border triple” that includes all solutions sets in the form (4k, 6k, k), so the garden border triple (4, 6, 1) includes (4, 6, 1), (8, 12, 2), (12, 18, 3), and so on (just like the Pythagorean triple (3, 4, 5) includes all solution sets in the form (3k, 4k, 5k)).

Are there other garden border triples other than (4, 6, 1)?  Just as there are different Pythagorean triple solutions to the formula a² + b² = c² ((3, 4, 5), (5, 12, 13), etc.), there are also different garden border triples.  And just as there are different Pythagorean triple generators (see here), there are different garden border triple generators.  To make one, we must generalize the garden border problem by calling the length of the inner garden b and the width of the inner garden h, as pictured below:

The length of the whole garden can then be expressed as 2x + b, the width of the whole garden can be expressed as 2x + h, and the area of the whole garden can be expressed as 2bh, giving us the equation (2x + b)(2x + h) = 2bh.  This time, however, we are going to solve this equation for b.  Multiplying (2x + b)(2x + h) gives us 4x² + 2bx + 2hx + bh = 2bh, subtracting bh on both sides gives us 4x² + 2bx + 2hx = bh, subtracting 2bx on both sides gives us 4x² + 2hx = bh – 2bx, factoring 2x from the left side and b from the right side gives us 2x(2x + h) = b(h – 2x), and dividing both sides by h – 2x gives us b = ^{2x(h + 2x)}/_{h – 2x}.

We can now use the formula b = ^{2x(h + 2x)}/_{h – 2x} to generate garden border triples.  If we let x = 1, then b = ^{2(h + 2)}/_{h – 2}.  Then if h = 3, b = ^{2(3 + 2)}/_{3 – 2} = 10, then the garden border triple is (3, 10, 1).  If h = 4, b = ^{2(4 + 2)}/_{4 – 2} = 6, then the garden border triple is (4, 6, 1) (which is a repeat of a triple we already knew).  If h = 5, b = ^{2(5 + 2)}/_{5 – 2} = ¹⁴/₃, then the garden border triple is (5, ¹⁴/₃, 1), and to eliminate the fraction we can multiply each number by 3 to get (15, 14, 3).  Continuing on in this fashion, we also arrive at (6, 4, 1) (a repeat), (7, ¹⁸/₅, 1) ≡ (35, 18, 5), (8, ¹⁰/₃, 1) ≡ (24, 10, 3), (9, ²²/₇, 1) ≡ (63, 22, 7), (10, 3, 1) (a repeat) and so on.  If we let x = 2, then b = ^{4(h + 4)}/_{h – 4}, and the resulting garden border triples are (5, 36, 2), (6, 20, 2) ≡ (3, 10, 1) (a repeat), (7, ⁴⁴/₃, 2) ≡ (21, 44, 6), (8, 12, 2) ≡ (4, 6, 1) (a repeat), (9, ⁵²/₅, 2) ≡ (45, 52, 10), (10, ²⁸/₃, 2) ≡ (15, 14, 3) (a repeat), and so on.  If we let x = 3, then b = ^{6(h + 6)}/_{h – 6}, and the resulting garden border triples are (7, 78, 3), (8, 42, 3), (9, 30, 3) ≡ (3, 10, 1) (a repeat), (10, 24, 3), and so on. 

The garden border problem is a common word problem given to students to practice solving quadratic equations.  Most solutions come out as a decimal answer, but there are a few scenarios in which the width, length, and border width are all integers, which we called garden border triples.  Generalizing the problem in terms of b and h and solving for b gave us a garden border triple generator b = ^{2x(h + 2x)}/_{h – 2x}.  The unique garden border triples we generated in this article were (3, 10, 1), (4, 6, 1), (15, 14, 3), (35, 18, 5), (24, 10, 3), (63, 22, 7), (5, 36, 2), (21, 44, 6), (45, 52, 10), (7, 78, 3), (8, 42, 3), and (10, 24, 3); but there are many, many more.