Dimensions Problem

Question

 A roll of chain-link fence is 800 ft ling.  What are the maximum dimensions of a rectangular field that can be enclosed with this fence?  What is the maximum area of this field?

Answer 1

The maximum area occurs when the rectangle is a square.

800/4=200, so you want a square 200 feet by 200 feet

Solving the problem algebraically...

P=2l+2w

800=2l+2w

400=l+w

400-l=w

A=lw

A=l(400-l)

A=400l-l²

the vertex of a parabola is (h,k) where h=-b/2a

the vertex is the maximum (or minimum) value of the parabola

h is the maximizing width and k is the maximum area

h=-b/2a

h=-400/(-2)

h=200 feet

You want a square that is 200 feet by 200 feet. You could plug 200 into A=400l-l² to get k, the maximum area but all you have to do is use A=s², A=200²=200*200=40,000 square feet for the maximum area

Answer 2

Courtney,

 

With these types of problems, I take a visual approach.

 

Try drawing a few different possible shapes of a four-cornered field enclosed by a fence. Try drawing one really narrow field that's long horizontally; try another that's narrow but vertically long. Think about how long the long sides are in those cases, and how long you could possibly make them. Try drawing some squatter fields, where the length and width are closer to being equal. Try drawing a square.

 

Now try finding the areas of those fields. Which are the biggest? Which are the smallest? Are the areas bigger when the fields are narrow or squat?

 

Now, I'm not sure if you're in a calculus course. If you are, you can use optimization techniques to maximize area.

 

You can create two equations from the information giving you, one for the perimeter of the field and one for the area.

 

Perimeter = the sum of the lengths of the four sides = the total length of your fencing, which is what?

 

And area = length * width.

 

From the equation for perimeter, you can solve for either length or width and plug into the equation for area. Then you'll have area in terms of only one variable. You can then take the derivative, set it equal to 0, and solve for the optimal length or width.

 

You can then plug that into the equation for perimeter to solve for the other one. Then you'll have length and width, and can solve for the optimum area.