SOLUTION: I need help!!! Suppose that you need to fence a rectangular play area in your backyard for your child or pet. Further, suppose that you know the length must be 8 feet longer than

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Question 892960: I need help!!!
Suppose that you need to fence a rectangular play area in your backyard for your child or pet. Further, suppose that you know the length must be 8 feet longer than the width. The back of your house will serve as one side of the fenced area. Note: The perimeter (distance around) of a general rectangle is P = 2L + 2W, and its area is A = L x W. In this situation, P = L + 2W. The value for the area is 4500.
2.Write the equation of the perimeter in terms of the length, L, only
P=2W+2L
Write the area equation in terms of the length, L, only.


What can you observe about the characteristics of that quadratic area function? Will this quadratic function’s graph cross the horizontal axis? How do you know?




Show all your work for finding both the length and the width of this rectangular fenced area.

Show all your work for calculating the cost of the fence.
Show all your work for calculating the cost per square foot of the fenced area.)
What observations and conclusions can you make about the results of them?)
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39623)   (Show Source): You can put this solution on YOUR website!
w for width
L for length
A for area
A=4500
L=8+w
Using your choice of 1*L to serve as one side of the rectangle region to be fenced, you choose L+2w for the length of fencing to use, c.
c for length of fencing to use.

Perimeter of the rectangle region is 2w+2L, but your length of fencing is c=L+2w.



------ One unknown variable, w, and one known variable, A as a given constant.
You have the choice to substitute for A now and if factorable quadratic, then factor the quadratic expression to solve for w; or keep going in symbolic form and substitute for A later. Seeing so many different factorizations for 4500, my work will be done first in symbolic form.


------- The PLUS form will be what is needed


, keeping like this for now.




If you rationalize the denominator and simplify (not showing those steps here),
----not simplifiable further.

The amount of fence material c=L+2w

You can then compute this as a decimal approximation if needed.
You could also first compute L and w and then find the value for c.
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!
I need help!!!
Suppose that you need to fence a rectangular play area in your backyard for your child or pet. Further, suppose that you know the length must be 8 feet longer than the width. The back of your house will serve as one side of the fenced area. Note: The perimeter (distance around) of a general rectangle is P = 2L + 2W, and its area is A = L x W. In this situation, P = L + 2W. The value for the area is 4500.
2.Write the equation of the perimeter in terms of the length, L, only
P=2W+2L
Write the area equation in terms of the length, L, only.


What can you observe about the characteristics of that quadratic area function? Will this quadratic function’s graph cross the horizontal axis? How do you know?




Show all your work for finding both the length and the width of this rectangular fenced area.

Show all your work for calculating the cost of the fence.
Show all your work for calculating the cost per square foot of the fenced area.)
What observations and conclusions can you make about the results of them?)


Length: L

Width:  L - 8

P, or perimeter of this enclosure: 2W + L, OR

P = 2(L - 8) + L

P = 2L - 16 + L





Area = LW

A = L(L - 8)

Area equation: , or 



Solve this using the quadratic equation formula, or by completing the square to determine L,

or length of enclosure. Then find its width. 

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