The Figure of Maximum Area and Given Perimeter
Date: 06/02/98 at 16:42:11 From: Brad Morris Subject: Calculus/geometry max/min problem Hi Math Doctor! I am a math teacher at Friends Central School in Wynnewood PA. In our Geometry book (USCMP) it states that the geometric figure of maximum area and given perimeter is a circle. In the margin it states that Calculus is required to prove it. I'm the Calculus teacher and our Geometry specialist asked me about it and I really couldn't come up with a proof. I need help! Thanks. Brad Morris
Date: 06/02/98 at 18:54:43 From: Doctor Anthony Subject: Re: Calculus/geometry max/min problem We can start by first showing that for, say, a rectangular shape of given perimeter, the square gives the greatest area. If the perimeter is P, let length be x, then width is (1/2)(P-2x) and area is: A = (1/2)x(P-2x) = Px/2 - x^2 dA/dx = P/2 - 2x = 0 for max or min So 2x = P/2 and x = P/4. Thus the square is the optimum shape for a 4-sided figure. This can be extended to figures with more sides, and it is always true that a symmetrical shape will enclose the greatest area for a given perimeter. If we have a circle with radius r the perimeter is 2*pi*r. If this is also the perimeter of a square, then each side of the square has length 2*pi*r/4 = pi*r/2 and area of square is: (pi*r/2)(pi*r/2) = (pi/4)pi*r^2 = pi/4 * area of circle and since pi/4 < 1 the area of the square is less than the area of the circle. A similar exercise with a 6-sided figure gives: area of hexagon = sqrt(3)pi/6 * area of circle = 0.9068 * area of circle If we then go to an n-sided figure with perimeter P, the area of one of the basic triangles formed by joining the midpoint of the figure to the vertices is: (1/2)r * cos(pi/n) * P/n (here r = distance from centre point to a vertex), and with n such triangles the area is (1/2)r * P * cos(pi/n). This will tend to its greatest value as cos(pi/n) -> 1 which requires that n -> infinity. So we end up with a polygon of an infinite number of sides, which is a circle. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 06/02/98 at 22:43:33 From: Doctor Pat Subject: Re: Calculus/geometry max/min problem Brad, This is not a calculus answer, but I think if I had to explain to a typical geometry student why the theorem was true, this would be my approach. STEP 1: Given any convex polygon with a given perimeter P, we can increase the area if any two adjacent sides are not the same. To illustrate this we draw a random n-gon with not all sides congruent. If two adjacent sides, call them r and s, are not congruent, draw the chord joining their two non-adjacent vertices, call it b, and form a triangle. The rest of the n-gon's area is not affected by changing the two sides r and s. We can improve the area if we maximize the triangle with a base equal to the chord b, and two sides that have a sum of r+s. I think most Geometry classes have proven that this triangle is maximized when the triangle is isosceles. Repeating this around the sides, eventually they must all end up the same. What we have demonstrated is that of all polygons with N sides, the regular polygon has the highest ratio for area/perimeter. STEP 2: Given any regular polygon with perimeter P and N sides, A regular Polygon with N+1 sides and perimeter P has a larger area. To illustrate this we begin with a regular n-gon of any number of sides. We know that any such n-gon is circumscribable so we draw the circle around it. Remember that the area of the n-gon is given by A = .5*Perimeter*apothem, which can be rewritten as A/P = apothem/2. That is, for any given perimeter, the area/perimeter ratio is directly proportional to the size of the apothem. We now show any one of the triangles formed by two radii of the circumscribing circle and one side of the n-gon. The perpendicular bisector of this side of the n-gon forms the altitude of the triangle, and the apothem of the n-gon. As we reduce the size of the side form P/n to P/(n+1) the angle gets a little smaller, and the ratio of the apothem to the radius of the circumscribing circle (which is a little smaller also for the n+1 gon) gets a little larger. We need to show that the apothem is actually larger than in the previous case by using one of the trig functions, probably the cosine. Thus we show that the perimeter has stayed constant, but the apothem has gotten bigger, indicating that the area is larger. I would expect this to be received much better if we first took say an octagon or decagon and found the area and perimeter and walked through the verification of each step. If your kids are like mine, they will nod 100 times when I show a proof they don't quite understand, but if I confirm it with numbers, they not only know it's true, they actually believe it's true, which is more important for me. Good luck. -Doctor Pat, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 11/07/2000 at 09:54:53 From: George Subject: Finding the largest area using the same perimeter How do I prove that the shape with the largest area and with a perimeter (or circumference) of 1000 m is a circle? Also, where do I go once I have done this? Is there an extension?
Date: 11/07/2000 at 15:11:00 From: Doctor Rob Subject: Re: Finding the largest area using the same perimeter Thanks for writing to Ask Dr. Math, George. Here is one way to proceed: Step 1: Show that a polygon with a fixed perimeter and largest area must be convex. Do this by considering any vertex with an interior angle of more than 180 degrees, and constructing the diagonal connecting the two vertices adjacent. The diagonal and the two edges will form a triangle. Reflect that triangle in the line that is the diagonal extended, and you have a polygon with a larger area and the same perimeter. Step 2: Show that a convex polygon with a fixed perimeter and largest area must have all sides of the same length. Do this by looking at two unequal adjacent sides, and the triangle formed by them and the diagonal connecting their two outside ends. Show that the isosceles triangle with the same base and perimeter has a larger area, so replacing the constructed triangle with the isosceles one will give a polygon with the same perimeter and a larger area. Step 3: Show that a convex polygon with equal sides, fixed perimeter, and largest area, must have all its interior angles of the same measure. Do this by looking at two adjacent angles of unequal measure, and the three sides forming them. Connect a diagonal from the outside ends of the sides to form a quadrilateral. Show that the isosceles trapezoid formed by the diagonal and three sides of the same length as the sides of the polygon has the same perimeter and larger area than the quadrilateral constructed, so replacing the three given sides and two included angles with three sides of the same lengths and two equal included angles forms a polygon with the same perimeter and larger area. (The fact that, given the lengths of the sides, the quadrilateral with the largest area is cyclic, that is, all its vertices lie on a single circle, should help here.) These three steps show that among all polygons with n sides and a fixed perimeter, the regular n-gon has the largest area. Step 4. Show that the area K of a regular n-gon with fixed perimeter P increases as n increases. Each side has length P/n, so K = n*(P/n)^2*cot(Pi/n)/4 K = (P^2/4*Pi)*(Pi/n)/tan(Pi/n) so you are reduced to showing that x/tan(x) increases as x approaches 0 from above (in fact, its limit is 1.) The limiting case of a regular n-gon as n increases without bound is a circle, which thus is the solution to the problem. There is another solution using the calculus of variations, which I won't elaborate on here. Can you generalize to polyhedra and spheres? - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Date: 04/08/2003 at 18:06:16 From: Mark Subject: Area of a triangle What are the dimensions of a triangle with perimeter p that encloses the maximum area?
Date: 04/08/2003 at 21:31:06 From: Doctor Paul Subject: Re: Area of a triangle The maximum area occurs when the triangle is equilateral with each side having length p/3. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/
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