IF THIS WAS 1500 SQFT. WHAT WOULD IT LOOK LIKE.I have two small Yorkies that love to go outside and play, so being on a leash just doesn’t work for them. So I decided to get them a fence so they can play how they like too. I live in a townhouse, its 1600 sqft yet its big enough for them. I will end up needing 165 ft of fencing to keep my dogs safe in my backyard.
2. L=2w+5 and w= 3L-4
3. A= LW
1600 sqft= (2W+5)w
1600 sqft= 2W^2 + 5W
2W^2+ 5W- 1600 sqft= 0
4. Next I used to the quadratic formula to figure out what w= which is x = +/- sqrt (b^2 – 4ac)/2
W= -5+/- sqrt (5^2-4(-1600)/4
W= -5+/- sqrt (25 + 6400)/4
W= (-5+/- 105)/4
W=( -5 + 105)/4
W+ (-5 – 105)/ 4
W= 20 ft and w= 27. 5
Length cannot be negative so therefore the w= 27.5
L= 2(27.5) + 5
L= 60 ft
5.) To find the perimeter the formula is P= 2L+2W
P= 2(60) + 2(27.5)
P=110+55
P= 165 ft
To determine what a 1500 sqft area would look like in terms of length and width for your backyard, you can follow these steps:
1. Start by representing the relationship between the length (L) and width (W) with the given equations:
L = 2W + 5 (Equation 1)
W = 3L - 4 (Equation 2)
2. Use the formula for the area (A), which is the product of length and width:
A = LW
3. Since you know the total area (1500 sqft), you can substitute the equations from step 1 into the area equation:
1500 sqft = (2W + 5)W
4. Simplify the equation by expanding and rearranging terms:
1500 sqft = 2W^2 + 5W
5. Now, you have a quadratic equation in terms of W. To solve for W, set the equation equal to zero:
2W^2 + 5W - 1500 sqft = 0
6. Apply the quadratic formula to find the values of W:
W = (-5 ± √(5^2 - 4 * 2 * -1500)) / (2 * 2)
W = (-5 ± √(25 + 12000)) / 4
W = (-5 ± √12025) / 4
7. Simplify the square root:
W = (-5 ± 109.6) / 4
8. Calculate both possible values for W:
W1 = (-5 + 109.6) / 4 = 26.4
W2 = (-5 - 109.6) / 4 = -28.65
Since width cannot be negative, discard W2.
9. Now that you have W, substitute it back into Equation 1 to find L:
L = 2W + 5 = 2(26.4) + 5 = 57.8
10. The dimensions for a 1500 sqft area would be approximately 57.8 ft for length and 26.4 ft for width.
Note: This explanation assumed that the dimensions are real numbers. If you are working with integers or a specific range of values, you may need to adjust the steps accordingly.