A man uses 60m of fencing to make 3sides of a rectangular fence,the fourth being the wall,if the area enclosed is 448m^2 .Find the possible length of the sides of the fence.
let the width be x m
let the length be y m
So we need 2 widths and 1 length
given: xy= 448
y = 448/x
2x + y = 60
2x + 448/x = 60
times x
2x^2 + 448 = 60x
x^2 - 30x + 224 = 0
(x-16)(x-14) = 0
x = 16 or x = 14
then y = 448/16 or y = 448/14
y = 28 or y = 32
The field could be 16 m wide and 28 m long, (the 28 m side parallel to the barn)
or
The fields could be 14 m wide and 32 m long, ( the 32 m side parallel to the barn)
Let's start by assuming the length of the rectangular fence is L and the width is W.
Since the fence has 3 sides, and each side uses 60m of fencing, the total length of the three sides is 3L.
The fourth side is the wall, so we don't need to account for the length of that side.
According to the problem, the total length of fencing used is 60m. Therefore, we can write the equation:
2L + W = 60m (Equation 1)
The area enclosed by the fence is given as 448m^2. The formula for area of a rectangle is:
Area = Length x Width
Substituting the given values, we have:
448m^2 = L x W
Now, we need to express the width (W) in terms of the length (L) to simplify the equation. We know that:
2L + W = 60m
Rearranging this equation, we get:
W = 60m - 2L
Substituting this value of W into the area equation, we have:
448m^2 = L x (60m - 2L)
Expanding this equation, we get:
448m^2 = 60L - 2L^2
Rearranging this equation, we get a quadratic equation:
2L^2 - 60L + 448 = 0
Now, we can solve this quadratic equation to find the possible values for L.
Factoring the above equation, we have:
2(L - 14)(L - 16) = 0
Setting each factor to zero, we get:
L - 14 = 0 or L - 16 = 0
Solving for L, we find two possible solutions:
L = 14 or L = 16
Substituting these values back into Equation 1 to find the corresponding width (W), we have:
When L = 14: 2L + W = 60
2(14) + W = 60
28 + W = 60
W = 60 - 28
W = 32
When L = 16: 2L + W = 60
2(16) + W = 60
32 + W = 60
W = 60 - 32
W = 28
Therefore, the possible lengths and widths of the fence are:
Length = 14m and Width = 32m
Length = 16m and Width = 28m
To find the possible lengths of the sides of the fence, we can start by setting up equations based on the given information.
Let's assume the length of the rectangular fence is "L" and the width is "W".
We know that the perimeter is equal to 60 meters, and the perimeter of a rectangle is calculated by adding the lengths of all four sides. Since we have 3 sides made of the fence material and the fourth side being the wall, we have:
2L + W = 60
We are also given that the area enclosed by the fence is 448 square meters. The area of a rectangle is calculated by multiplying its length by its width, so we have:
L * W = 448
Now we have a system of two equations with two variables. We can solve this system by substitution or elimination:
1. Solving by substitution:
Rearrange the first equation to solve for W:
W = 60 - 2L
Substitute this value of W into the second equation:
L * (60 - 2L) = 448
60L - 2L^2 = 448
2L^2 - 60L + 448 = 0
Now, we can solve this quadratic equation to find the possible values of L. Since we have a quadratic equation, we can use the quadratic formula:
L = (-b ± sqrt(b^2 - 4ac)) / 2a
Using the values of a = 2, b = -60, and c = 448, we can substitute them into the formula and solve for L. After finding the values of L, we can substitute them back into the equation W = 60 - 2L to find the corresponding values of W.
2. Solving by elimination:
Multiply the first equation by 2 to make the coefficients of W equal:
2(2L + W) = 2(60)
4L + 2W = 120
Now we have:
4L + 2W = 120
L * W = 448
By substituting the value of W from the first equation to the second equation, we can eliminate the variable W and solve for L.
Once we find the possible values of L and W, we need to check if they are reasonable and positive. In this case, since we are dealing with dimensions, both L and W must be positive. Additionally, the values of L and W should satisfy the conditions of the problem.
By solving the system of equations, we can find the possible lengths of the sides of the fence.