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# GMAT Math: Learn from Nature

Written by Kelly Granson. Posted in GMAT Study Guide

What does a cat do when it prepares for sleep on a cold night? It tucks in its paws and rolls up its body to make itself as round as possible. It does so, obviously, to keep itself warm, to minimize release of heat through the body surface. Given its volume, which cannot be decreased, the cat is minimizing its surface area. In other words, the animal solves the problem of the body with the smallest surface area for a given volume, making itself more spherical. This cat appears to know something about isoperimetric problems! And you, if you plan to take the GMAT, should familiarize yourself with them as well.

Historical Background

A straight line is the shortest distance between two points. The arc of a circle is the shortest curve that can connect two points on a sphere. Among all closed plane curves of the same length, the largest area is covered by a circle, and among all closed surfaces of the same area the largest volume is enclosed in a sphere.

Maximum and minimum properties such as these were known to the ancient Greek mathematicians, even before the proofs were available.

Virgil (Publius Vergilius Maro), one of the most famous poets of ancient Rome, wrote The Aeneid based on a widespread legend, apparently related to the events of the Ninth Century BC. Phoenician princess Dido, fleeing from persecution, after many adventures arrived at the coast of Africa (now the Gulf of Tunis), where she founded the city of Carthage and became its first queen. For her fortress, she purchased a plot of land, "not more than can be surrounded by an oxhide." Then she cut the oxhide into narrow strips, which she tied together into a long rope. Now she faced a mathematical problem: in what shape should that rope be laid out to enclose a piece of the greatest possible area? In memory of this story, the Carthaginian fortress was called "Byrsa," which in the language of the inhabitants of Carthage means "oxhide."

The exact mathematical formulation of this problem is as follows: Among closed plane curves of a given length, find a curve that encloses the maximum area. This problem is called a Dido problem or isoperimetric problem: isoperimetric figures are figures that have same-sized perimeters. Many historians believe that this is the first extremal problem ever to be discussed in the scientific literature. Virgil, describing this legend, used the verb "circumdare" (surround), which contains the root "circus" (circle), suggesting that Dido solved the problem correctly, i.e. that the plot had to have a circular perimeter.

Rules for Isoperimetric Problems

1. For a given length of line, the largest area is enclosed by a circle.

2. Of all triangles with given perimeter, the equilateral triangle has the largest area.

3. Of all triangles with two given sides a and b, the largest in area will be a right triangle with legs a and b.

4. Of all right triangles with a given hypotenuse, the isosceles right triangle has the maximum area.

5. Of all rectangles with the given perimeter, the square has the maximum area.

6. Of all the bodies with a given surface area, the sphere has maximum volume.

Make sure you memorize these six rules.

Look at the following four examples that demonstrate how isoperimetric concepts may be incorporated into GMAT questions.

Example 1: For a given length of line, the largest area is enclosed by a circle.

If Bob was provided with a rope of length 12π feet, then what is the maximum area that can be enclosed by this rope?

(A) 9 feet2

(B) 9π feet2

(C) 36π feet2

(D) 64π feet2

(E) 144π feet2

Solution

For any curve of a given length, the circle formed by that curve covers the greatest possible area. Therefore, you need to find the area of a circle with a circumference of 12π feet. Recall that Circumference =Diameter × π. Hence, 12π feet=Diameter × π, from which you can calculate that the diameter is 12 feet, so the radius must be 6 feet.

The area of the circle must be π(radius)= π(6 feet)= 36π feet2.

Example 2: Of all right triangles with a given hypotenuse, the isosceles right triangle has the maximum area.

In rectangle JKLM shown above, is the area of triangle JKR greater than 17?

1. JR=8

2. JK=3

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient.

Solution

Statement (1):

In any right triangle with a given hypotenuse, the maximum area is attained when the two sides containing the right angle are equal in length. In other words, the maximum area for any right triangle with a given hypotenuse is attained when that right triangle is an isosceles right triangle.

If right triangle JKR is isosceles and the hypotenuse of right triangle JKR is 8, then the length of each of the two equal sides of triangle JKR would be , as triangle JKR is a 45-45-90 triangle. Then, the maximum area that triangle JKR would attain is . Therefore, no right triangle with a hypotenuse of 8 could attain an area greater than 16, so the area of triangle JKR cannot be more than 17.

The answer to the question is NO.

Statement (1) ALONE is sufficient.

Statement (2):

You know that JK=3 but you do not know the length of KR. For instance, if KR=2, then the area of triangle JKR would be , while if KR=18, for instance, then the area of triangle JKR would be , for example. As a result, without additional information you cannot determine whether the area of triangle JKR is more than 17.

Statement (2) ALONE is NOT sufficient.

Example 3: Of all the rectangles with a given perimeter, the square has the maximum area.

John Joe needs to fence the village he wants to build for his fellow ranchers with fencing material of length 84 yards. If John Joe wants to build a rectangular village with the greatest possible area for his fellow ranchers, then which of the following would be the best choice of dimensions for his village?

(A) 41 yards × 1 yard

(B) 11 yards × 31 yards

(C) 21 yards × 11 yards

(D) 21 yards × 21 yards

(E) 20 yards × 21 yards

Solution

For any rectangular shape with a given perimeter, the maximum area is attained when the rectangle takes the shape of a square. If the perimeter of the rectangular village must be 84 yards (since John Joe only has 84 yards of fencing materials for fencing his village), then a square with sides of length  yards each would have a perimeter of 84 yards.

Therefore, the maximum area for this village would be attained when the dimensions of the village are 21 yards × 21 yards.

You could also solve this problem by reverse engineering. If you try each answer choice to determine which one yields the greatest area, then you could answer the question provided that the given dimensions for the village in the answer choice produces a perimeter of 84 yards for the village.

Example 4: Of all triangles with two given sides a and b, the largest area will be that of a right triangle with legs a and b.

If Horace Horton's only cow and only horse are pegged to the same stake on a patch of open space in Horace Horton's ranch by two separate ropes of length 70 feet each, then what is the maximum area of the triangle that can be formed by joining the peg, the cow, and the horse by straight lines?

(A) 1225 feet2

(B) 2450 feet2

(C) 2450π feet2

(D) 4900 feet2

(E) 4900π feet2

Solution

You need to understand the question first. Both the cow and the horse are tied by ropes of length 70 feet each to a peg, or stake, hammered into the ground. If the cow moves around the peg with a taut rope (where the rope is stretched to its full length), it will describe a circular path around the peg, and the horse will do the same.

Essentially, you are asked to determine the maximum area that can be formed by the triangle formed by joining the peg, the horse and the cow by straight lines, where the positions of the horse and the cow are on the circumference of the circle, and the peg/stake is the centre of the circle. This circle would have a radius of 70 feet.

In simpler terms, you need to find out the maximum area of any triangle that can be formed by joining the centre with two points on the circumference of a circle with radius 70 feet.

For any triangle with two known sides, the maximum area is attained when the angle between those two sides is a right angle. Then, the maximum area for a triangle with two sides of length 70 feet each will be , and this area will be attained when the two sides are perpendicular to each other.