Question

Mike's family wants to build a rectangular fenced backyard area for their dog. They have a 20-meter length of wire fence and four posts. They can also use the 20-meter straight length of the back of their house as a side of the enclosure, but the fence cannot attach directly to the house. The fence must stretch taught between posts, and they have fasteners to attach the fence to the posts. Describe or sketch a design for the dog enclosure that yields the maximum area using the resources they have. List the dimensions of the enclosure and list the total area. Check your work by describing a similar design that does not enclose as much area.
I need help figuring out how to set up this up. Thanks!

Answer 1

Answer:

• See the image showing the design attached and the explanation below.

• Dimensions of the maximum enclosure:

        - two sides of 5m

       - one side of 10 m

• Check: see the check below.

       

Explanation:

1. Design

The image attached shows the design of the dog enclosure: there are four posts, two sides of fence run perpendicular to the back of the house, and the third side runs parallel to the back of the house. Thus the three sides of the fence are  stretched taught between post using fasteners.The back of the house is the fourth side of the enclosure area.

The two sides of the fence that are perpendicular to the back of the house have the same length and are labeled x.

The other side is labeled 20 - 2x. Thus the total length is

20 - 2x + x + x = 20, to represent the 20-meter length of wire fence.

2. Maximum area

The enclosure area has shape of rectangle. Thus, the area is:

         $Area=length\times width\\\\Area=(20-2x)x\\\\Area=20x-2x^2$

Since the expression $20x-2x^2$ represents a parabola that opens downward, the vertex is the maximum value of the function.

Then, find the vertex for the equation $y=20x-2x^2$

Complete squares to find the vertex form of such parabola:

         $y=20x-2x^2\\\\y=-2x^2+20x\\\\y=-2(x^2-10x)\\\\y=-2(x^2-10x+25)+50\\\\y=-2(x-5)^2+50$

Comparing with the general vertex form the vertex is (5, 50).

It means that the maximum value of the area is 50m², when the x = 5.

Therefore, the sides of the enclosure are:

• x = 5m

• 20m - 2x = 20m - 2(5m) = 20m - 10m = 10m

Hence, the the dimensions of the enclosure are two sides of 5m and one side 10 m.

3. Check your work by describing a similar design that does not enclose as much area.

Choose, for example, x = 6 .

$Area=(6m)\times(20m -2(6m))=6m\times(20m-12m)=6m\times 8m=48m^2$

You can take any other value of x and you will obtain, always, an area less than 50m² confirming that 50m² is the maximum area.