Question

Suppose you had to use exactly 200 m of fencing to make either one square enclosure or two separate square enclosures of any size you wished. What plan would give you the least area? What plan would give you the greatest area?

Answer 1

Answer:

Greatest area: One square enclosure with side 50 m

Least area: Two square enclosures with side 25 m each

Step-by-step explanation:

We know we have 200 m of fencing to make the required enclosures. Since the fence will run surrounding any enclosure, it is called the perimeter.

The perimeter for any square area of side z is computed as

P=4z

And its area is

$A=z^2$

Now, let's analyze both options (shown in the figure below)

Option 1: One square enclosure

Knowing P=200 m, we can determine the length of the side

z=200 m /4 = 50 m

The area is easily computed

$A=50^2=2500 m^2$

Option 2: Two separate square enclosures of any size

Let's say the side of one of them is x and the side of the other one is y

Assuming both enclosures have no sides in common, the total perimeter is

P=4x+4y

We have 200 m to make the job, so

4x+4y=200

Or equivalently

x+y=50 => y=50-x

The total area of both enclosures is

$A=x^2+y^2$

Replacing the expression of y

$A=x^2+(50-x)^2$

To know what the best value is for x to maximize or minimize the area, we use derivatives with respect to x

$A'=2x+2(50-x)(-1)$

$A'=2x-100+2x=4x-100$

We equate A'=0 to find the critical point

4x-100=0

x=25 m

Since y=50-x

y=25 m

And the total area is

$A=25^2+25^2=1250\ m^2$

Note: if we set any other combination for x and y, say x=20 m and y=30m we would get greater areas

$A=20^2+30^2=1300\ m^2$

The first option gives us the greatest area of 2500 m^2 and the second option has the least area of 1250 m^2