Shapes, EfficientEighteen ounces that Salubrious grain is packaged in a box (inches) × × 11". The box"s volume is around 180 cubic inches; its total surface area, A, is around 249 square inches. Could the manufacturer keep the same volume but reduce A by transforming the dimensions of the package? If so, the agency could conserve money.

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Consider 3" × 6" × 10" and also 4" × 5" × 9" boxes. Their volume is 180 cubic inches, however A = 216 and also 202 square inches, respectively. Clearly, because that a solved volume, surface area deserve to vary, suggesting that a certain shape might produce a minimum surface area.

The trouble can be rephrased this way: let x, y, and z it is in the political parties of a box such that xyz = V for a solved V. Find the minimum worth of the function A (x,y,z) = 2xy + 2xz + 2yz. This looks complicated. An appropriate strategy would certainly be to first solve a much easier two-dimensional trouble in order to develop methods because that attacking the three-dimensional problem.

Consider the following: let the area that a rectangular region be fixed at 100 square inches. Discover the minimum perimeter the the region. Letting x and also y it is in the sides of the rectangle method that xy = 100, and you room asked to minimize the perimeter, p (x, y ) = 2x + 2y. Begin by solving the first equation for y so the the 2nd can be expressed as a role of one variable, specific . The difficulty can currently be solved in number of ways. First, graph the function, obtaining the graph below.

Using the techniques obtainable on the graphing calculator, you can find the the minimum role value occurs at x = 10, do y = 10, and also P = 40.

Consider yet one more simplification to our "minimum area" problem— permit the basic of the box be a square, thereby eliminating one variable. If the basic is x through x and also the elevation is y, climate a fixed V = x 2y and a variable A (x, y ) = 2x 2 + 4xy. Due to the fact that . Making use of calculus, , offering The optimal equipment is a cube.

Calculus provides it clear that the minimum surface area occurs as soon as the grain box is a cube, which method the area amounts to 6V 2/3. In the original problem, a cube through edges equal to ≈ 5.65 provides a surface ar area of roughly 192 square inches, a savings of 23% from the original box. It have to be noted that nonrectangular shapes may reduce the surface ar area further. Because that example, a cylinder v a radius that 2.7 inches and also a volume that 180 cubic inch would have actually a full surface area the 179 square inches. A sphere, while not a feasible form for a cereal box, would have a surface area the 154 square inch if the volume is 180 cubic inches. The larger question then becomes: that all 3 dimensional shapes with a addressed volume, which has the least surface area?

History of the effective Shape Problem

Efficient shapes problems emerged as soon as someone asked the question: the all airplane figures through the very same perimeter, which has the greatest area? Aristotle (384 b.c.e.–322 b.c.e.) may have actually known this problem, having noted that a geometer knows why ring wounds cure the slowest. Zenodorus (c. 180 b.c.e.) created a treatise on isometric figures and also included the complying with propositions, not all of which were confirmed completely.

Of all constant polygons through equal perimeter, the one v the most sides has the biggest area.A circle has a better area than any kind of regular polygon that the exact same perimeter.A sphere has a greater volume 보다 solid numbers with the same surface area.

Since Zenodorus color etc on the occupational of Archimedes (287 b.c.e.–212 b.c.e.), it is possible that Archimedes contributed to the equipment of the problem. Zenodorus supplied both direct and indirect proofs in combination with inequalities including sides and also angles in order to establish inequalities amongst areas that triangles. Pappus (c. 320 c.e.) prolonged Zenodorus"s work, including the proposition that of every circular segments through the exact same circumference, the semicircle stop the biggest area. In a memorable passage, Pappus wrote that v instinct, living creatures might seek optimal solutions. In particular he was reasoning of love husband bees, noting that they made decision a hexagon because that the form of honeycombs—a wise an option since hexagons will certainly hold an ext honey for the very same amount of material than squares or it is intended triangles.

After Pappus, difficulties such as these go not attract successful attention until the development of calculus in the so late 1600s. If calculus created tools for the solution of countless optimization problems, calculus was not able come prove the to organize that claims that the all plane figures through a solved perimeter, the circle has the greatest area. Making use of methods drawn from synthetic geometry, Jacob Steiner (1796–1863) progressed the proof considerably. Schwarz, in 1884, ultimately proved the theorem. Research study on efficient shapes is still flourishing.

One important element of these difficulties is that they come in pairs. Paired through the trouble of minimizing the perimeter the a rectangle of addressed area is the dual problem of maximizing the area the a rectangle with fixed perimeter. Because that example, the equipments of a cereal box may want to have actually a big area because that advertising. Similarly, paired through the problem of minimizing the surface ar area that a form of resolved volume is the trouble of maximizing the volume the a form of addressed surface area. Typical calculus problems involve detect the size of a cylinder that minimizes surface ar area for a fixed volume or that pair. The cylinder who height amounts to its diameter has the the very least surface area because that a addressed volume.

see also Minimum surface ar Area; Dimensional Relationships.

Don Barry


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

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

Heath, sir Thomas. A background of Greek Mathematics. New York: Dover Publications, 1981.

Kazarinoff, Nicholas D. Geometric Inequalities. Brand-new York: L. W. Singer, 1961.

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Knorr, Wilbur Richard. The ancient Tradition of Geometric Problems. Brand-new York: Dover Publications, 1993.