2w+2l = 50
w+l=25 or l = 25-w
Also wl = 150
w(25-w) = 150
25w - w^2 - 150 = 0
w^2 - 25w + 150 = 0
solve the quadratic using your favourite method, however I can plainly see nice factors.
w+l=25 or l = 25-w
Also wl = 150
w(25-w) = 150
25w - w^2 - 150 = 0
w^2 - 25w + 150 = 0
solve the quadratic using your favourite method, however I can plainly see nice factors.
Let's assume the length of the play yard is L and the width is W.
The equation for the perimeter (P) of a rectangle is given by:
P = 2L + 2W
Given that Tim needs to install 50 feet of fencing, we have:
P = 50
To find the dimensions, we need to consider the relationship between the perimeter and the area of the rectangle.
The area (A) of a rectangle is given by:
A = L * W
Given that the area of the play yard is 150 square feet, we have:
A = 150
Now, we have a system of equations:
2L + 2W = 50 (equation 1)
L * W = 150 (equation 2)
To solve this system of equations, we can use substitution or elimination method.
Let's solve it using the substitution method:
1. Rearrange equation 1 to express L in terms of W:
2L = 50 - 2W
L = 25 - W
2. Substitute the value of L in equation 2:
(25 - W) * W = 150
3. Simplify and rearrange the equation:
25W - W^2 = 150
W^2 - 25W + 150 = 0
4. Factor the quadratic equation:
(W - 10)(W - 15) = 0
Setting each factor equal to zero gives us:
W - 10 = 0 or W - 15 = 0
Solving for W:
If W - 10 = 0, then W = 10
If W - 15 = 0, then W = 15
Now, we have two possible values for the width of the play yard: W = 10 or W = 15.
Substituting these values back into equation 1 to find the corresponding lengths:
If W = 10, then L = 25 - W = 25 - 10 = 15
If W = 15, then L = 25 - W = 25 - 15 = 10
Therefore, the dimensions of the play yard could be either 15 feet by 10 feet or 10 feet by 15 feet.