Optimization?

Question

A large rectangular yard is to be laid out and fenced in, and divided into 4
enclosures by fences parallel to one side of the yard
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If 2 miles of fencing is available, what is the maximum area (in square miles)?

Answer 1

There are two strategies for dividing the rectangle, one where 3 inside fences are parallel to one side and the other where 2 internal fences are each parallel to one side or the other.

Case1: 3 parallel internal fences
Case2: 2 crossing internal fences

Assume the rectangles sides are x and y. Then the area of the outside rectangle is A = x*y. Furthermore:

For case 1:
5x + 2y = 2 miles
y = (2 - 5x)/2
A = x*y = x(2-5x)/2
dA/dx = (2 - 10x)/2
To maximise are set dA/dx = 0 -> 2-10x = 0 -> x = 1/5
for x = 1/5 y = (2 - 5*1/5 )/2= 1/2
Max Area = 1/5 * 1/2 = 1/10 sq mile

and for case 2:
3(x+y) = 2 miles
y = 2/3 - x
A = x(2/3-x)
dA/dx = 2/3 - 2x -> 2/3 - 2x = 0 -> x = 1/3 -> y = 2/3 -1/3 = 1/3
Max Area = 1/3 * 1/3 = 1/9 sq mile

Note the max area for case 2 is larger where there are 2 crossing internal fences. -> Max Area = 1/9 sq Mile

/m

Answer 2

OK, assume the dimensions of the yard are x by y. The value to be maximized is the area, given by:
A = xy

The amount of fencing used to enclose the yard is given by
2x + 5y = 2.

Solve for y in terms of x:
y = (2-2x)/5

Substitute back into the area equation:
A = (2x-2x²)/5 = -(2/5)x² + (2/5)x

Take the derivative of A:
A' = -(4/5)x + 2/5

Solve for x when A' = 0:
0 = -(4/5)x + 2/5
x = 0.5

At this value, we have either a local minimum or a local maximum. To find out which, take the second derivative of A:
A' = -(4/5)x + 2/5
A'' = -4/5

Since A'' is negative, we have a maximum.

Plug this back into the equation for A to find the enclosed area:
A = -(2/5)x² + (2/5)x
A = -(2/5)(1/4) + (2/5)(1/2) = 0.1

So the total area enclosed is 0.1 square miles.

(Edit: perplexed*: Good answer, but the wording of the problem requires that the space is divided by fences parallel to ONE side of the yard.)

Answer 3

Let L = a length of fence to make the width of one enclosure
Let H = a length of fence to make the height of one enclosure

So we want 8L + 5H = 2 miles such that we maximize 4L * H

8L + 5H = 2
8L = 2 - 5H
L = (2-5H)/8

4L * H = max
4(2-5H)/8 * H = max
(2-5H)H/2 = max
2H - 5H^2 = max

max' = 2 - 10H = 0
2 = 10H
H = 1/5 mile

8L + 5H = 2
8L + 5(1/5) = 2
8L = 1
L = 1/8 mile

So the maximum area would be:

4L * H = 4(1/8) * 1/5 = 1/10 square miles

(Edit: jgoulden, you would have to use more than 2 miles of fencing with the dimensions you gave... don't forget that the fencing must compose 4 enclosures, so there are 13 total lengths of fencing)

Answer 4

Let the length be 4L and the width be W.

Total linear fence = 2 miles = 2 ( 4L ) + 5 ( W ) = 8L + 5W
so W = ( 2 - 8L ) / 5

Area = 4 L W = 4 L ( 2 - 8L ) / 5 = ( 8L - 32L² ) / 5

To maximize, take the derivative and set equal to zero

0 = 8 / 5 - 32 L / 5
32L = 8
L = 1 / 4

then W = 1 - 4 L = 1 - 4 ( 1 / 4 ) = 1 - 1 / 4 = 3 / 4 mile

The area is then ( 3 / 4 ) ( 4 ) ( 1 / 4 ) = 3 / 4 square mile