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Maximize Area with Two Subdivisions (Non-Rectangular)

Updated on October 25, 2016
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TR Smith is a product designer and former teacher who uses math in her work every day.

A typical calculus optimization problem is to find the optimal dimensions of a rectangular enclosure built with a fixed length of fence, subdivided into two equal rectangular sections that share a border. This is something you may encounter in real life when sectioning off part of your land to build a garden or animal pen.

A more difficult (yet equally real-world applicable) problem is to find the find the dimensions of the enclosure without the restriction that it or its subdivisions be rectangular. This opens up infinitely many possibilities for the general shape of the fenced enclosure. In this calculus tutorial we will solve this more general problem, but restrict ourselves to only four possible shapes, which are shown in the image above: rectangle, circle, stadium, and rhombus. In the real world, these shapes are easy to construct with fencing material. Shapes such as parabolas, ellipses, and regular polygons may be more difficult to construct with precision.

If you want to skip the complicated math and go directly to the solution, click here to go to the bottom of the page.

Case 1: Classic Rectangular Enclosure

This problem may be familiar to many calculus students since it is the one usually presented in textbooks.

Suppose Kathy has 30 feet of fencing material and wants to build a rectangular enclosure divided into two equal-area sections by a line of fence down the middle. What should the dimensions of the enclosure be to maximize the total area?

In the diagram above, if the total width is of the enclosure is W feet and the total length is L feet, then 3W + 2L = 30. This can be written equivalently as L = 15 - 1.5W.

Since the area is the width times length, we have

Area = WL
= W(15 - 1.5W)
= 15W - 1.5W^2

To find the value of W and L that yield the maximum area, we take the derivative of the area function, set it equal to zero, solve for W, and then back solve for L. This gives you

Area Derivative = 15 - 3W
15 - 3W = 0
15 = 3W
W = 5 feet

L = 15 - 1.5W
= 15 - 1.5*5
= 15 - 7.5
= 7.5 feet

Therefore, the maximum area occurs when the width is 5 feet and the length is 7.5 feet. The total area in this case is 5 times 7.5 equals 37.5 square feet. So how does this result compare to the area that can be obtained with other shapes?

Case 2: Circular Enclosure Divide by Its Diameter

This case requires no calculus because there is only one variable, D, the diameter of the circle. If the enclosure and its dividing line of fence has the shape in the figure above, and the total length of fencing material is 30 feet, then we have 30 = (π+1)D, or equivalently. D = 30/(π+1).

Since the area of a circle is (π/4)D^2, the total area of the enclosure is

(π/4)*[30/(π+1)]^2 ≈ 41.2095 square feet.

This is an improvement over the optimal rectangular fenced enclosure. With this construction the area increases by 9.89%.

Case 3: Rhombus with Divider Connecting Opposite Corners

Suppose Kathy has 30 feet of fence and builds the subdivided enclosure in the shape of a rhombus with two equal sized triangular sections. The four sides of a rhombus have equal length, call it T. If the acute angle of the rhombus is θ then the length of the dividing line is 2Tsin(θ/2). Therfore, the total perimeter is 4T + 2Tsin(θ/2). See diagram above.

Starting with the perimeter equation 30 = 4T + 2Tsin(θ/2), we can solve this equation to isolate T, which gives us T = 15/(2 + sin(θ/2)).

The area of a rhombus with a side length of T and acute angle of θ is given by the equation

Area = (T^2)sin(θ)

Using the identity T = 15/(2 + sin(θ/2)), we can write the area formula solely in terms of θ. This gives us

Area = 225sin(θ) / [2 + sin(θ/2)]^2

To maximize the area, we need to take the derivative of this equation with respect to θ, set the expression equal to 0, and solve for θ. This will give us the angle that yields the maximum area. To find the correct side length, we plug that value of θ into the relation T = 15/(2 + sin(θ/2)).

A(θ) = 225sin(θ) / [2 + sin(θ/2)]^2
A'(θ) = [450cos(θ) - 225sin(θ/2)] / [2 + sin(θ/2)]^2
0 = [450cos(θ) - 225sin(θ/2)] / [2 + sin(θ/2)]^2
2cos(θ) = sin(θ/2)

In radians, the exact solution is

θ = 4*arctan[ 1/4 + sqrt(33)/4 + sqrt(18 + sqrt(132))/4 ] radians

In degrees, this is approximately 72.75 degrees. Back-solving for the side length gives T ≈ 5.7846 feet.

The total area of the enclosure with this shape is 31.9571 square feet, which makes this the worst of the shapes analyzed so far.

Case 4: Stadium-Shaped Enclosure

A stadium is a oval-like shape formed by taking a rectangle and capping opposite sides with semi-circles. In the diagram above, we consider a stadium shape whose semi-circlular ends have a diameter of Y feet and whose straight sides have a length of X feet. If the enclosure is divided down the middle with a line of fence as shown above, and Kathy is still restricted to 30 feet of fence material, then we have

30 = πY + Y + 2X
= (π+1)Y + 2X

Solving this equation for X gives us X = 15 - (π + 1)Y/2. The area of the stadium is the combined area of the two semi-circles plus the rectangular piece in the middle. This works out to

Area = (π/4)Y^2 + XY
= (π/4)Y^2 + (15 - (π + 1)Y/2)Y
= 15Y - (π/4 + 1/2)Y^2

Taking the derivative with respect to Y and setting equal to zero gives us

Area' = 15 - (π/2 + 1)Y
0 = 15 - (π/2 + 1)Y
(π/2 + 1)Y = 15
Y = 30/(π + 2) feet

Back-solving for X gives us

X = 15 - (π+1)*30/(2π + 4)
X = 15/(π + 2) feet

Plugging the optimal values of X and Y back into the area equations gives us

Area = 225/(π + 2)
43.7608 square feet.

Compared to the other three shapes, this one provides the largest area of the enclosure. Therefore, to maximize the area Kathy should build the subdivided enclosure in the shape of a stadium.


It turns out that bending the fence into the shape of a stadium yields the maximum area out of all the four possibilities. If the fixed amount of fence is L feet in length, then the diameter of the semi-circular caps must be L/(π+2) ≈ 0.19449*L, and the length of the straight segments must be L/(2π + 4) ≈ 0.097246*L. The total area of the enclosure is (L^2)/(4π + 8).

The runner-up is the circle. The third-place solution is the classic rectangular case. Finally, the rhombus yields the least area of the four.


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