SOLUTION: Construction. A garden area is 30 ft. long and 20 ft. wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft^2, what is the width of the pat

Question 107596: Construction. A garden area is 30 ft. long and 20 ft. wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft^2, what is the width of the path?
Boy do I ever need help with this one. Thank you!
Found 2 solutions by scott8148, bucky:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
let x=width ... (30-2x)(20-2x)=400 ... 600-100x+4x^2=400 ... 200-100x+4x^2=0

x^2-25x+50=0 ... use quadratic formula to find x
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
You have an area of 20 ft by 30 ft in which to build the walkway and the garden. This means
that the available area for the garden and walkway is to be 20 times 30 or 600 square feet.
Since the garden itself is to be 400 square feet, that means that the remaining area for the
walkway is 600 sq ft minus 400 sq ft = 200 sq ft.
.
Here comes the hard part ... picturing the walkway. Call the width of the walkway W.
.
Now take a piece of notebook paper and arrange it so that its length runs left and right,
and its shorter dimension runs up and down. Draw a border (like a picture frame) around the
edges of the paper. (Make the border uniform all around the paper and of width about an inch.
This is just a sketch so don't waste time making the drawing really accurate.)
.
Your paper should look something like this:
.
|////////////////////////////////////////| ^
|////////////////////////////////////////| |
|////........................................................................////| |
|////........................................................................////| |
|////........................................................................////| |
|////........................................................................////| |
|////........................................................................////| 20 ft
|////........................................................................////| |
|////////////////////////////////////////| |
|////////////////////////////////////////| |
.
<------------------30 ft ---------------------------->
.
I hope this figure looks OK, but it may not when you get it. Hopefully, you'll get the idea.
.
In this sketch the slant bars represent the walkway and the dots represent the garden plot.
.
Look at the walkway (first 2 rows of slant bars) along the top edge of the paper). From
the left edge to the right edge it represents 30 ft long. Its width (two rows of slant bars) is
W units wide. So the entire area of the first two rows of slant bars is 30W. Notice at
the bottom of the sketch there is also an area of walkway that is two rows of slant bars
running the "30 ft length of the paper". Therefore, its area is also 30W. Combined, these
two parts of the walkway have a total area of 30W + 30W which equals 60W.
.
Now let's look at the areas of the walkway along the 20 ft sides. NOTICE that the length
of the unaccounted for walkway area along the 20 ft side is 20 ft less the two W widths at the
corners which have already been taken care of in the 30 ft length of the long side. Therefore,
the length of the walkway along the 20 ft edge is 20 ft - 2W and this length gets multiplied
by the width of the walkway (W) to give you an area of (20 - 2W)*W which multiplies out
to an area of 20W - 2W^2. And there are two areas of this size ... one along the left 20 ft side
and one along the right 20 ft side. Therefore, the total walkway area along both 20 ft sides
is 2*(20W - 2W^2) = 40W - 4W^2.
.
The total area of the walkway is the sum of the area of the 20 ft sides plus the area of the walkway
along the 30 ft sides. That is:
.
60W + 40W - 4W^2
.
and above we determined that this area was to be 200 square ft. So we have derived the
equation we need as being:
.
60W + 40W - 4W^2 = 200
.
Combine the 60W and the 40W and the equation is:
.
-4W^2 + 100W = 200
.
Subtract 200 from both sides and you have the standard quadratic form of:
.
-4W^2 + 100W - 200 = 0
.
Just to make the form a little more conventional, let's multiply both sides (all terms) by -1
to get:
.
4W^2 - 100W + 200 = 0
.
4 is a common factor of all the terms, so we can divide all terms on both sides by 4 to get:
.
W^2 - 25W + 50 = 0
.
This is of the standard quadratic form aW^2 + bW + C = 0 in which a = 1, b = -25, and C = 50
.
The quadratic formula says that the answer to such an equation is:
.

.
You can then solve for W by substituting the appropriate values for a, b, and C to get:
.

.
Inside the radical (-25)^2 = 625 and -4*1*50 = -200. Substituting these values results in:
.

.
and since the square root of 425 is 20.6155 the problem reduces to:
.

.
and dividing both terms in the numerator by the denominator 2 results in:
.

.
So there are two possible answers for W. One is 12.5 + 10.3078 = 22.3078. However, we
can ignore this answer. This would make the width of the walkway bigger than the 20 ft
dimension of one side and therefore it is an impossible solution.
.
The other solution for W is:
.

.
This answer for W (that is W = 2.1922 ft) looks good. Let's try it:
.
30 ft * 2.1922 + 30ft * 2.1922 = 65.766 + 65.766 = 131.532 sq ft
.
So the area of the two portions of the walkway that are 30 ft long is 131.532 sq ft.
.
Next, the unaccounted for walkway along the 20 ft side is (20 - 2(2.1922))ft long by 2.1922
ft wide ... and there are two of these sections. If you run these out on a calculator
you get that each section is 15.6156 ft long by 2.1922 wide for an area of 34.2325 sq ft.
So the area of the two sections is 2 times 34.2325 = 68.465 sq ft.
.
Adding 131.532 sq ft and 68.465 sq ft results in 199.997 sq ft of walkway area. We had figured
that the walkway area was to be 200 sq ft, and this is close enough. Therefore, we can say that
the width of 2.1922 ft for the walkway is the answer to this problem.
.
Hope this helps. It's getting late so you'd better check my logic and all my figures. Never
trust a sleep deprived tutor ... LOL.

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