How do you find maximum area?

Question

In geometry how do you find maximum area if, say, the given perimeter is 36. I'm sure it's really easy I just totally forgot. Thanks!

Answer 1

Here we use the Area-Perimeter relationship. Let's say we have a rectangle with length a, width b, perimeter P and area is A. Then we have:
A = ab
P = 2a + 2b

Now we let P be constant. Here a and b are variables (meaning the dimensions of the rectangle varies). Thus,
a = (P - 2b)/2

We substitute this to the area:
A = (P - 2b)/2 · b

We simplify:
A = Pb/2 - b²

Or,
A = -b² + P/2 b

Now here comes the calculus part... we use the tool called "differentiation" to find the "slope of the tangent line". When the slope of the tangent line is 0, then the value is either relatively maximum or minimum. The symbol for "slope for tangent line" is dA/db. When you differentiate, you will get
dA/db = -2b + P/2

Since we want the tangent line to be zero, then
0 = -2b + P/2

Solving for b, we have
b = P/4

solving for a,
a = (P - 2b)/2 = (P - P/2)/2 = (P/2)/2

Thus,
a = P/4

Thus, the maximum area is when
a = P/4 and b = P/4.

We notice that a = b. This is a very common rule. Just remember, "To get the maximum product, we set the factors be equal."

^_^_^_^_^_^

Answer 2

Finding Maximum Area

Answer 3

Actually, it's not so easy to figure out, but the answer is easy once you do. Basically, you take an n-sided polygon and figure out its area based on the side length, which is known to be (36/n). Then you can look at a graph of that equation and see where the area maximizes.

It's a pain to do, and really frustrating because the result is so simple. The maximum area for a given perimeter is a circle. Here's a real-life example that I've never forgotten: the balloon. When you blow up a balloon, you are increasing its volume. But the rubber has a tendency to shrink when you pull on it. What ends up happening is that the rubber is going to shrink to the smallest size that it can while holding in all the air you've put into the balloon. That's why balloons are round.

The maximum area enclosed by a 36-foot perimeter is then:
C = 2pi r
r = C/2pi

A = pi r^2 (substitute for r from the previous equation)
A = pi (C/2pi)^2
pi is 3.14 and C is 36 so...
A = 103.18 square feet

Of course, I just re-read the question and realized you may have forgotten to mention something, such as you're limited to 4 sides or something. That would be a very important detail.

Answer 4

[Give it to DaNNiX. The boy's a genius.] ______ [METHOD 1] Let the rectangle width be 2x and height y. Let the perpendicular height of the isosceles triangle be z. Then by Pythagoras the side s of the triangle is: s = √(x²+z²) The perimeter and area of the pentagon are thus: p = P(x,y,z) = 2x + 2y + 2√(x²+z²) = 2x [ 1 + y/x + √(1+(z/x)²) ] A(x,y,z) = 2xy +2*(1/2)x√(x²+z²) = 2x² [y/x + √(1+(z/x)²) ] and we want to maximize R(x,y,z) = A(x,y,z) / P(x,y,z) = 2x² [y/x + √(1+(z/x)²) ] / 2x [ 1 + y/x + √(1+(z/x)²) ] Writing y/x = Y , z/x = Z Extracting the common subterm u = y/x + √(1+(z/x)²) R(x,u) = which is maximized when (∂R/∂x, ∂R/∂u) = (0,0) = ( u/(1+u), x/(1+u)² ) and that says u=0 but also x=0 (??) => u = Y + √(1+Z²) = 0 => Y is negative (??) Seems to suggest it hs no extrema? _________________________________ [METHOD 2] Let's start by getting a baseline result ignoring the triangle. Let rectangle have aspect ratio 2a, thus let its width be 2a and height 1. Add the triangle of height z. Sidelength s = √(a²+z²) = a√(1+Z²) with Z=z/a Perimeter P(a,z) = 2 + 2a + 2a√(1+Z²) Area A(a,z) = 1*2a + a*a√(1+Z²) = a[2 + a√(1+Z²)] Then ratio: R(a,z) = A(a,z)/P(a,z) = 2[1 + a + a√(1+Z²)] / a[2 + a√(1+Z²)] Write v = √(1+Z²) R(a,v) = 2(1+a+av) / a(2+av) which is maximized when (∂R/∂a, ∂R/∂v) = (0,0) ∂R/∂a = [ (2a+a²v)2(1+v) - 2(1+a+av)(2+2av) ] / a²(2+av)² = 2[ 2a +2av +a²v +a²v² -2 -2av -2a -2a²v -2av -2a²v² ] / a²(2+av)² = 2[ 2 +a²v +2av +a²v² ] / a²(2+av)² ∂R/∂v = [ a(2+av)(2a) - 2(1+a+av)a² ] / a²(2+av)² = 2a²[ (2+av) - (1+a+av) ] / a²(2+av)² = 2(1-a) / (2+av)² Well ∂R/∂v has an extremum @ a=1 and ∂R/∂a = 0 => (2 +a²v +2av +a²v²) = 0 substituting a=1: v² +3v +2 = 0 v = -1 or -2 which is meaningless. Seems wrong. Maybe someone can point out my mistake.

Answer 5

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a = height of rectangle b = wiidth θ = angle in triangle Perimeter p = 2a + b + bsecθ = constant Area A = ab + 1/4 b^2 tanθ Now a = 1/2 * [p - b - bsecθ] A = 1/2 * [pb - b^2 - b^2 secθ] + 1/4 b^2 tanθ Take partial derivatives A_b = 1/2 * [p - 2b - 2bsecθ] + 1/2 b tanθ = 0 ... (1) A_θ = 1/2 (-b^2 secθtanθ) + 1/4 b^2 sec^2 θ = 0 This gives sinθ = 1/2 or θ = 30 deg Sub in (1) to get b = p * (2 - √3) a = p * (3 - √3) / 6 Finally A = p^2 (2 - √3) / 4

Answer 6

You can discover what shape surrounds the most area for the least perimeter by making a 36" long loop of string, and try arranging it in different shapes until you see which shape is the largest.

You will discover a single shape that maximizes area for a given perimeter. This shape is found a lot in nature, where an object can survive most efficiently by enclosing the most space with the least packaging.

Good luck

Answer 7

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RE:
How do you find maximum area?
In geometry how do you find maximum area if, say, the given perimeter is 36. I'm sure it's really easy I just totally forgot. Thanks!

Answer 8

Well if you are talking about Perimeter for a circle, you need to find the radius, since Perimeter of a circle = circumference of a circle.

P = 36 = 2*pi*r

r = 5.732

Area = pi r*r = 3.14*5.732*5.732 =103.18

Answer 9

perimetere of circle =2*pi*r
36=2*pi*r
r=18/pi

thus max area=pi*r^2

=pi*(18/pi)^2

=324/pi unit square