Struggling with a word problem

There doesn't seem to be enough information to get an answer on this one. I have looked around and been unable to find any example problems anywhere near this one. This is a 5-part question however I am assuming I will be able to find for the entire problem if I can get past this first part.

Suppose you wanted to construct a fence around a garden plot in the form of a rectangle. On the neighbor’s side it’s going to need heavy-duty fencing that costs $2.00 per foot. The other three sides can be made of standard fencing material that costs $1.20 foot. You have $200 to spend.

Write an equation using two variables for the total cost of the fence. Use x and y to represent the width and length of the rectangle. Solve the equation for one of the variables (either x or y). How many different rectangles would it be possible to enclose for $200?

All of the example problems similar to this one give the perimeter of the fence/plot which would make this easier to tackle, but I am at a loss on this one. Any help?

Hello, Crookshanks!

Read the question again.
They did <u>not</u> ask for maximum area or minimum cost.

They asked you set up the equation . . . There is no "answer".

Suppose you wanted to construct a fence around a garden plot in the form of a rectangle.
On the neighbor’s side it’s going to need heavy-duty fencing that costs $2.00 per foot.
The other three sides can be made of standard fencing material that costs $1.20 foot.
You have $200 to spend.

Write an equation using two variables for the total cost of the fence.
Use x and y to represent the width and length of the rectangle.
Solve the equation for one of the variables (either x or y).
How many different rectangles would it be possible to enclose for $200?

Click to expand...

The front and sides requires 2x + y feet of fencing.
. . . At $1.20 per foot, those sections will cost: .1.2(2x + y) dollars.

The back border requires y feet of fencing.
. . . At $2.00 per foot, that section will cost: .2y dollars.

Hence the total cost is: . 1.2(2x + y) + 2y .= .2.4x + 3.2y dollars.

The budget is limited to $200, so: . 2.4x + 3.2y .= .200

Solving for y, we get: . y .= .(-3/4)x + (125/2)


"How many different rectangles . . . ?"
A strange question . . . There is an <u>infinite</u> number of possible rectangles.

The x's can range from 0 feet up to 83 1/3 feet.
The corresponding y's range from 62 1/2 feet down to 0 feet.

I understand better about not reading the question correctly. Thanks for the help. Could you elaborate a little though on how you determined that "The x's can range from 0 feet up to 83 1/3 feet.
The corresponding y's range from 62 1/2 feet down to 0 feet." ? Thanks. I was doing so well with the synthetic division in this chapter then I hit this question and am so lost.

Hello, Crookshanks!

Could you elaborate a little though on how you determined that
"The x's can range from 0 feet up to 83 1/3 feet. The corresponding y's range from 62 1/2 feet down to 0 feet." ?

Click to expand...

I threw that in to show how much variation there could be.

I took that equation: .2.4x + 3.2y = 200 and then I went for the "extremes":
. . . Let x = 0, then y = 125/2
. . . Let y = 0, then x = 250/3