In some problems, the GRE may require you to determine the maximum or minimum area of a given figure. Following a few simple shortcuts can help you solve certain problems quickly.
Perhaps the best-known maximum area problem is one that asks you to maximize the area of a quadrilateral (usually a rectangle) with a fixed perimeter. If a quadrilateral has a fixed perimeter, say, 36 inches, it can take a variety of shapes:
Of these figures, the one with the largest area is the square. This is a general rule: Of all quadrilaterals with a given perimeter, the square has the largest area. This is true even in cases involving non-integer lengths. For instance, of all quadrilaterals with a perimeter of 25 feet, the one with the largest area is a square with 25 ÷ 4 = 6.25 feet per side.
This principle can also be turned around to yield the following corollary: Of all quadrilaterals with a given area, the square has the minimum perimeter.
Both of these principles can be generalized for polygons with n sides: A regular polygon with all sides equal and all angles equal will maximize area for a given perimeter and minimize perimeter for a given area.
Another common optimization problem involves maximizing the area of a triangle or parallelogram with given side lengths.
For instance, there are many triangles with two sides 3 and 4 units long. Imagine that the two sides of lengths 3 and 4 are on a hinge. The third side can have various lengths:
There are many corresponding parallelograms with two sides 3 and 4 units long:
The area of a triangle is given by
,
and the area of a parallelogram is given by A = bh. Because both of these formulas involve the perpendicular height h, the maximum area of each figure is achieved when the 3-unit side is perpendicular to the 4-unit side, so that the height is 3 units. All the other figures have lesser heights. (Note that in this case, the triangle of maximum area is the famous 3–4–5 right triangle.) If the sides are not perpendicular, then the figure is squished, so to speak.
The general rule is this: If you are given two sides of a triangle or parallelogram, you can maximize the area by placing those two sides PERPENDICULAR to each other.