Tim Brown wants to enclose his children's play area with fencing. Tim found that he needs to install 50 feet of fencing around the rectangular play yard. If the yard covers 150 square feet, what are its dimensions?
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.
To find the dimensions of the rectangular play yard, we can set up an equation based on the given information.
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.