# Shapes, Efficient

# Shapes, Efficient

Eighteen ounces of *Salubrious Cereal* is packaged in a box (inches) × × 11". The box's volume is about 180 cubic inches; its total surface area, *A*, is about 249 square inches. Could the manufacturer keep the same volume but reduce *A* by changing the dimensions of the package? If so, the company could save money.

Consider 3" × 6" × 10" and 4" × 5" × 9" boxes. Their volume is 180 cubic inches, but *A* = 216 and 202 square inches, respectively. Clearly, for a fixed volume, surface area can vary, suggesting that a certain shape might produce a minimum surface area.

The problem can be rephrased this way: Let *x*, *y*, and *z* be the sides of a box such that *xyz = V* for a fixed *V*. Find the minimum value of the function *A* (x,y,z) = 2xy + 2xz + 2yz. This looks complicated. An appropriate strategy would be to first solve a simpler two-dimensional problem in order to develop methods for attacking the three-dimensional problem.

Consider the following: Let the area of a rectangular region be fixed at 100 square inches. Find the minimum perimeter of the region. Letting *x* and *y* be the sides of the rectangle means that *xy* = 100, and you are asked to minimize the perimeter, *P* (*x*, *y* ) = 2*x* + 2*y*. Begin by solving the first equation for *y* so that the second can be expressed as a function of one variable, namely . The problem can now be solved in several ways. First, graph the function, obtaining the graph below.

Using the techniques available on the graphing calculator, you could find that the minimum function value occurs at *x* = 10, making *y* = 10, and *P* = 40.

Consider yet another simplification to our "minimum area" problem— let the base of the box be a square, thereby eliminating one variable. If the base is *x* by *x* and the height is *y*, then a fixed *V = x* ^{2}*y* and a variable *A* (*x*, *y* ) = 2*x* ^{2} + 4*xy*. Since . Using calculus, , giving The optimal solution is a cube.

Calculus makes it clear that the minimum surface area occurs when the cereal box is a cube, which means the area equals 6*V* ^{2/3}. In the original problem, a cube with edges equal to ≈ 5.65 gives a surface area of approximately 192 square inches, a savings of 23% from the original box. It should be noted that nonrectangular shapes may reduce the surface area further. For example, a cylinder with a radius of 2.7 inches and a volume of 180 cubic inches would have a total surface area of 179 square inches. A sphere, while not a feasible shape for a cereal box, would have a surface area of 154 square inches if the volume is 180 cubic inches. The larger question then becomes: Of all three dimensional shapes with a fixed volume, which has the least surface area?

## History of the Efficient Shape Problem

Efficient shapes problems emerged when someone asked the question: Of all plane figures with the same perimeter, which has the greatest area? Aristotle (384 b.c.e.–322 b.c.e.) may have known this problem, having noted that a geometer knows why round wounds heal the slowest. Zenodorus (c. 180 b.c.e.) wrote a treatise on isometric figures and included the following propositions, not all of which were proved completely.

- Of all regular polygons with equal perimeter, the one with the most sides has the greatest area.
- A circle has a greater area than any regular polygon of the same perimeter.
- A sphere has a greater volume than solid figures with the same surface area.

Since Zenodorus draws on the work of Archimedes (287 b.c.e.–212 b.c.e.), it is possible that Archimedes contributed to the solution of the problem. Zenodorus used both direct and indirect proofs in combination with inequalities involving sides and angles in order to establish inequalities among areas of triangles. Pappus (c. 320 c.e.) extended Zenodorus's work, adding the proposition that of all circular segments with the same circumference, the semicircle holds the greatest area. In a memorable passage, Pappus wrote that through instinct, living creatures may seek optimal solutions. In particular he was thinking of honey bees, noting that they chose a hexagon for the shape of honeycombs—a wise choice since hexagons will hold more honey for the same amount of material than squares or equilateral triangles.

After Pappus, problems such as these did not attract successful attention until the advent of calculus in the late 1600s. While calculus created tools for the solution of many optimization problems, calculus was not able to prove the theorem that states that of all plane figures with a fixed perimeter, the circle has the greatest area. Using methods drawn from synthetic geometry, Jacob Steiner (1796–1863) advanced the proof considerably. Schwarz, in 1884, finally proved the theorem. Research on efficient shapes is still flourishing.

One important aspect of these problems is that they come in pairs. Paired with the problem of minimizing the perimeter of a rectangle of fixed area is the dual problem of maximizing the area of a rectangle with fixed perimeter. For example, the makers of a cereal box may want to have a large area for advertising. Similarly, paired with the problem of minimizing the surface area of a shape of fixed volume is the problem of maximizing the volume of a shape of fixed surface area. Standard calculus problems involve finding the dimensions of a cylinder that minimizes surface area for a fixed volume or its pair. The cylinder whose height equals its diameter has the least surface area for a fixed volume.

see also Minimum Surface Area; Dimensional Relationships.

*Don Barry*

## Bibliography

Beckenbach, Edwin and Richard Bellman. *An Introduction to Inequalities.* New York: L. W. Singer, 1961.

Chang, Gengzhe and Thomas W. Sederberg. *Over and Over Again.* Washington, D.C.: Mathematical Association of America, 1997.

Heath, Sir Thomas. *A History of Greek Mathematics.* New York: Dover Publications, 1981.

Kazarinoff, Nicholas D. *Geometric Inequalities.* New York: L. W. Singer, 1961.

Knorr, Wilbur Richard. *The Ancient Tradition of Geometric Problems.* New York: Dover Publications, 1993.

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