Write a function to represent the area as the function of the width...?

Question

Situation: Owners of a farm want to fence in an area for goats. They want to use up 40 meters of fencing they purchase.

1) What is the maximum area they can provide for the animals with 40 meters of fence if they want the pen to be a quadrilateral?


2) How does changing the width affect the area?


3) Write a function to represent the area as the function of the width?



Bonus: Now imagine that the fenced-in area is adjacent to a building. You only need to fence three sides.



Solution would be highly appreciated.

Answer 1

In order to answer this question, it's best to do #3 first.

If the farmers have only 40m of fence, then the perimeter of the area will be 40m. The formula for the perimeter of a rectangle is p = 2l + 2w, so 40 = 2l + 2w, or dividing both sides by 2, 20 = l + w. You can also solve this for l: l = 20 - w.

Then since the area is length times width, A = lw, you can substitute (20 - w) from the first equation to get a formula for A in terms of w only:

A(w) = (20 - w)w = 20w - w²

Now, if you're familiar with parabolas, this is a parabola. The vertex (h, k) is the maximum value of this parabola:

y = (x - h)² + k

To put A(w) in this format, complete the square:

A(w) = -w² + 20w = -(w² - 20w)
= -(w² - 20w + 100) + 100
= -(w - 10)² + 100

So the vertex is (10, 100). This means that the maximum area occurs when x = 10 and that maximum area would be 100.

Also, looking at the equation in the parabola form, you should be able to see that any value of w other than 10 will reduce the value of the area.

So the answers again are:

1) 100 square meters.
2) When the width is between 0 and 10 meters, increasing the width increases the area, to a maximum of 100 m² when the width is 10m. Between 10 and 20 meters, increasing the width decreases the area.
3) A(w) = 20w - w²

If the fenced-in area is adjacent to a building, the perimeter equation would be p = w + 2l, so 40 = w + 2l, so l = 20 - 1/2 w.

Then A(w) = (20 - 1/2w)w = 20w - 1/2w².

Finding the vertex for this parabola:

A(w) = -1/2(w² - 40w)
= -1/2(w² - 40w) - 200 + 200
= -1/2(w² - 40w + 400) + 200
= -1/2(w - 20)² + 200

So the vertex here is (20, 200); therefore the area is at its maximum (200 m²) when the width is 20.

Answer 2

Ah, an optimization problem. First, your perimeter must be 40 m as specified in the problem. Let the length of one side be x the adjacent side be y. Then 2x + 2y = 40. Now, you want to maximize the area, which is given by A = xy. Solve the first equation for x or y, and plug into the area equation. Then you'll have an equation for the area with only one variable. Take the derivative of A with respect to x (or y), set that equal to zero and solve for x (or y). Then use the perimeter equation to find the other variable, multiply the two together and you have the maximum area. Conceptually, it's not difficult to reason that a square has the most area of any quadrilateral.