I need to find an area of 3 rectangles.

Question

Question: A breeder of horses wants to fence three rectangular grazing areas along a river with 600 meters of fencing. On one of the length sides there is a river and no fence is needed along that size. write an expression for the total area A of the grazing land in terms of the width W of the rectangles. 

What is the largest area she can enclose? 

 

      W

   ''''''''''''''''

   '          '

   '''''''''''''''

L '           '  River (no fence needed on this side) 

   '''''''''''''''

   '          '

   '''''''''''''''

   

 

Answer 1

Based on your picture, the length of the fence is L+4W, which you set equal to 600 meters and solve for L:

 

L+4W=600

L=600-4W

 

The total area is

 

A=LW=(600-4W)W=600W-4W²

 

You can get the maximum value of this quadratic function in one of three ways:

 

1. Using calculus: set the derivative equal to zero;

2. Graphing the function: parabola whose maximum is the vertex;

3. Using algebra: complete the square.

 

I will use method 3:

 

A= 600W-4W² = -(2W-150)²+150²

 

Since -(2W-150)² is always nonpositive, the maximum is reached when -(2W-150)² is zero, which implies

2W-150=0

W=150/2=75

A= 150²=22,500 m²

Answer 2

Hi Dexter,

 

The easiest way to do a problem like this is to make a figure. It is awesome that you already have one. The next step is to find what information is important, and how to start making that information into equations.

 

So you a) have 600 m of fencing and b) want to find the total area c) in terms of width.

 

With a) you have to think about what characteristic of the rectangle that the fencing acts as. It acts as the sides of the rectangle, becoming the perimeter. So the total fencing perimeter adds up to  600 m.

 

P = 600 m

 

So what exactly is the total perimeter? Add it up from the figure. I count 4 W and 3L. So...

 

P= 4W + 3L

     or

4W + 3L = 600

 

For b) you are calculating area of fenced region. Generally, for one rectangle you start with this equation:

 

Area = L * W

 

But you want the total area; there is not one rectangle. You have three. So the new equation is the area of three rectangles:

 

Area = 3 L * W

 

Lastly you have c). In the equation for area you have L and W; you need to find a way to have only only W in the equation. This is accomplished by substitution. You need to find what L is in terms of W. So that where L is in the equation, you substitute in what it is in W. I'll show you.

 

You have two equations. You want to keep the second equation, because the question is asking about area. Which is what the answer is asking about. So you use the first equation.

 

4W + 3L = 600                 // you are looking to find what L is equal to in terms of W

-4W       = -4W                // the first step is to "isolate the variable"

        3L = 600-4W

     %3  = %3 (600-4W)   //you want just L, not 3L so you divide by 3

        L  =  1/3(600 - 4W)

 

Now that you have an equation for L in terms of W you can substitute it into the second equation.

 

Area = 3 L * W

Area = 3 * 1/3(600-4W) * W

Area = (600-4W)*W                        // 3 * 1/3 = 1

 

To find the maximum you find the zeroes (where W can be set to make Area = 0) in the equation, and find the mean of those values. If it will help, graph the equation: It is a downward facing parabola-you can see the maximum in the center.

 

In this equation the zeroes are:

 

600-4W = 0                // if one term is made zero, then both terms by multiplication will become zero

600 = 4W                   //  Area = 0 * W = 0

150 = W

 

and

 

W = 0

 

Finding the mean of these values is to find the average.

 

Average = (Term1 + Term2)/(Number of terms)

Average = (150 + 0)/2

Average = 150/2 = 75.

 

Therefore, the largest area can be found with a width that is 75 m.

 

What is this area? Plug it in:

 

Area = (600-4W)*W

Area = (600-4(75))*75

Area = (600-300)*75

Area = 300*75

Area = 22500 m^2

 

This is the answer, congratulations!