Question

A group of students want to build three identical rectangular gardens, each of

which has a width of x ft. The gardens will be arranged in one of two ways, as
showr by the plans below. The students have 480 ft of fencing and plan to build
a fenc around each of the three gardens. They will also install lighting for the
garders.
Plan A
Plan B
X
x
X
X
X
1. Suppose that the students use all of the fencing. Given that the width of a
small rectangular garden is x ft, write an expression in terms of x for the length
of one of the small rectangular gardens in Plan A. Then write an expression
in terms of x for the length of one of the small rectangular gardens in Plan B.
Explain.

Answer 1

So, here we have the following figures for each plan:

We're given that the students have 480 ft of fencing, and they want to build a fence around each of the three gardens.

If "x" represents the width, we could say that "y" represents the large.

To find an expression for the length of one of the small rectangular gardens in plan A, remember that the perimeter of a rectangle is defined as the sum of the measures of all its sides. So, the total fence in the plan A is given by the expression:

`4x+6y=480`

We could solve this equation in terms of x to find the value of y:

`\begin{gathered} 6y=480-4x \\ y=(480-4x)/(6)\to y=80-(2)/(3)x \end{gathered}`

Now, we're asked to find an expression for the length of one of the small rectangular gardens in plan A.

So, the perimeter of one rectangle will be the substraction between 480 and the perimeter of the other two small rectangles:

`\begin{gathered} 480-(x+x+y+y+y+y) \\ 480-2x-4y \end{gathered}`

But we know that y = 80 - 2/3x, so:

`\begin{gathered} L=480-2x-4(80-(2)/(3)x) \\ L(x)=480-2x-320+(2)/(3)x \\ L(x)=160-(4)/(3)x \end{gathered}`

So that's the expression for the length of one of the small rectangular gardens in plan A.

Now, for plan B, we got that the total length of the three gardens is,

`5x+5y=480`

If we solve for y:

`\begin{gathered} 5y=480-5x \\ y=96-x \end{gathered}`

And, the perimeter of one rectangle will be the substraction between 480 and the perimeter of the other two small rectangles:

`\begin{gathered} L=480-(3x+3y) \\ L=480-3x-3y \end{gathered}`

But, we know that y=96-x, so:

`\begin{gathered} L(x)=480-3x-3(96-x) \\ L(x)=480-3x-288+3x \\ L(x)=192 \end{gathered}`

That means that the value of the length in the plan B is constant.