Max Area of Rectangle with 3 Sections and Fixed Fence Length
In the intersection of geometry, calculus, and optimization, a common problem is to find the maximum area of a region with a fixed perimeter. In the case of a region that consists of a single section where you can't use any existing walls, the maximum area is obtained by making the enclosure in the shape of a circle. If the region must be in the shape of a rectangle, then the maximum area is obtained by bending the given perimeter of fence into the shape of a square.
A more complicated problem is to find the maximum area when the region is subdivided into two or more sections that share a common boundary line. Here we solve the case of finding largest rectangular area with three subdivisions using a fixed length of fence material.
The 3-Section Rectangular Area Optimization Problem
Kay has 48 feet of fencing material and wants to build three contiguous rectancular enclosures that together form a larger rectangle. Each of the three sections must have the same area and adjacent sections will share fence. She has two possible configurations to choose from, shown in the image below.
To find the dimensions that provide the maximum area, we need to analyze each configuration separately.
The first configuration shows three equal sized rectangles in a row forming a larger rectangle with a width of W and a length of L. Since the total amount of fence is fixed at 48 feet, we have
48 = 2L + 4W
Area = L*W
Solving the first equation for L gives us L = 24 - 2W. Plugging this into the area equation gives us
Area = (24 - 2W)W
= 24W - 2W^2.
The derivative of this equation with respect to the variable W is
Area' = 24 - 4W
Setting the derivative equal to 0 and solving for W gives us
0 = 24 - 4W
W = 6 feet
Plugging this value of W into the perimeter equation gives us L = 12 feet. The total area of all three rectangular sections is 6*12 = 72 square feet, and the area of each 6-by-4 section is 24 square feet.
Now let's look a the case where each rectangular section is adjacent to the other two and all three have the same area. If the overall width and length of the outer rectangle are W and L respectively, and the total amount of fencing material is 48 feet long, then we have
48 = 3L + 2 ⅔ W = 3L + (8/3)W
Area = W*L
Solving the perimeter equation for L gives us L = 16 - (8/9)W. Plugging this into the area formula gives us
Area = W(16 - (8/9)W)
= 16W - (8/9)W^2
Taking the derivative of the area equation with respect to the variable W gives us
Area' = 16 - (16/9)W
And setting the derivative equal to zero and solving for W gives us
0 = 16 - (16/9)W
W = 9 feet
Plugging W = 9 into the perimeter equation gives us L = 8 feet. This means that the total area of the enclosure is 8*9 = 72 square feet. Two the the rectangular sections have dimensions of 4-by-6 feet, and one of the sections is 3-by-8 feet. Each section has an area of 24 square feet.
Notice that both configurations have a maximum total area of 72 square feet and that each configuration yields sections of 24 square feet. Moreover, each configuration has 8 vertices. Therefore, either configuration yields the maximum area with a fixed length of fence.
Non-Rectangular Solutions to the Area Optimization Problem
Kay restricted herself to constructing rectangular enclosures with her limit of 48 feet of fencing material. However, there are other arrangements that yield larger areas but involve shapes other than rectangles. For example, a stadium-oval shape divided into two semi circles and one rectangle, or a circle divided into three sectors, provide larger areas with the same 48 feet of fence.
The stadium shape above with its dividing lines uses a total of 48 feet of fence. It has a total area of 77.2672 square feet, and each section has an area of 25.7557 square feet. This is a 7.3% improvement over the optimal rectangular solutions.
The circle shape below with its dividing lines also uses a total of 48 feet of fence. It has a total area of 83.9921 square feet, and each section has an area of 27.9974 square feet. This is a 16.7% increase over the optimal rectangular solutions.