A farmer has a rectangular garden plot surrounded by 200 ft of fence. If its area is 1600 ft2, what is the length and width of the garden?

Answer 1

Perimeter = P = 200

Area = 1600

A = Lw, P = 2L+2w

Lw =1600

w = 1600/L

200 = 2L + 2(1600/L)

100 = L + 1600/L

L^2 - 100L + 1600 = 0

L^2 -100L +2500 = 2500 -1600 = 900

(L-50)^2 = 30^2

L -50 = + or - 30

L = 20 or 80 feet Long

w = 1600/L = 80 or 20 feet wide

or just factor L^2 -100L +1600 = (L-80)(L+20) and set each factor = 0 and solve for L=20 and 80

Answer 2

Hi,

I hope you are doing well! This sort of question will best be answered by devising a systems of equations and solving for both the length and the width.

Because this is a rectangle, we know that the area is going to be equal to the length times the width.

A = width * length

For simplification purposes, I will assign x to represent the width and y to represent the length.

Since we know that the area is 1600, we can rewrite the equation to look like this:

1600 = xy

Now, imagine a rectangular fence, the perimeter is going to be equal to 2 times the width + 2 times the length.

Perimeter = 2*width + 2*length

Perimeter = 2W + 2L

200 = 2X + 2Y

To be it simpler, I am going to simplify

100 = X + Y

Now, we have these 2 equations:

xy = 1600

x + y = 100

Now, I am going to isolate x from the perimeter equation

x + y = 100

x = 100 - y

We can now substitute this equation for the width in the area equation.

xy = 1600

(100 - y)y = 1600

100y - y^2 = 1600

Now I am going to rearrange this to make it easier to isolate y.

100y = 1600 + y^2

0 = y^2 - 100y + 1600

*We can factor this equation into the following:

(y - 80)(y-20) = 0

Now the length can be either 80 or 20 feet

Now let's solve for x or the width

1) y is 80 feet

x = 100 - y

x = 100 - 80 feet

x = 20 feet

or

2) y is 20 feet

x = 100 - y

x = 100 - 20 feet

x = 80 feet

So now the width can either be 20 or 80 feet. These combinations look familiar, right?

How do we know which one is which?

Well we don't but we do know that there are 2 possible scenarios as listed:

Scenario #1: The length is 80 feet and the width is 20 feet

Scenario #2: The width is 20 feet and the length is 80 feet

Just for a good measure, you can also substitute these values back into the original equations just to make sure that they work!

Hope that helps!

Answer 3

2x + 2y = 200

x y = 1600

2x = 200 - 2y

x = 100 - y

(100-y)y = 1600

100y - y^2 = 1600

0 = y^2 - 100y + 1600

0 = (y - 80)(y - 20)

y = 80 or 20

x = 100-y = 100-80 or 100-20 = 20 or 80

Length = 80

Width = 20