Liz has a rectangular garden with a walkway around the edge. with the walkway included the entire garden area has a 20 foot length and an 18 foot width. At the edge of the walkway it is twice as wide as it is at the edge of the width. The area of the garden itself without the walkways 200 ft.². How wide is each part of the walkway?
L=20-4x
W=16-2x
A= (20-4x)(16-2x)
A=320-40x-64x+8x²
A=8x²-104x+320
I used the quadratic formula and got x=15.57 ft or x=-2.57 ft.
So the width of the short side is 15.57 ft and the width of the long side is 31.14 ft
I see so many things wrong here.
Why are you taking 16-2x, when according to your question it should be 18-2x
which is it?
You also completely ignored the fact that the whole garden has an area of 200 f^2
Where did it say that the area of the walkway is zero?
That is what you are saying by solving it
using the quadratic equation formula
(your x= 15.57) ???
To solve this problem, we can follow these steps:
1. Let's define the dimensions of the rectangular garden. We'll call the length L and the width W. We are given that the total area of the garden with the walkway is 20 ft by 18 ft.
2. We are also told that the width of the walkway on the long side is twice the width at the edge of the width. This is given by the equation W = 2x, where x is the width at the edge of the width.
3. The length of the garden with the walkway is given by L = (20 - 2x), since we need to subtract the width of the walkway on both sides. Similarly, the width of the garden with the walkway is given by W = (18 - 4x) because we need to subtract the width of the walkway on each side.
4. The area of the garden without the walkway is given as 200 ft². We can calculate this area by multiplying the length and width of the garden itself: (20 - 4x) * (18 - 2x) = 200.
5. Now, let's expand and simplify the equation: (20 - 4x) * (18 - 2x) = 200. Multiplying the terms gives us: 360 - 40x - 36x + 8x² = 200.
6. Rearranging the equation, we get: 8x² - 76x + 160 = 0.
7. We can now solve this quadratic equation to find the value of x, which represents the width of the walkway. You mentioned that you used the quadratic formula, and the solutions are x = 15.57 ft and x = -2.57 ft.
8. However, since the width of the walkway cannot be negative, we can discard the negative solution. Therefore, the width of the walkway is approximately 15.57 ft.
9. Finally, since the width at the edge of the width is twice the width of the walkway, we can calculate it by substituting the value of x in W = 2x. Thus, the width at the edge of the width is approximately 31.14 ft.
To summarize, the width of the walkway is approximately 15.57 ft, and the width at the edge of the width is approximately 31.14 ft.