1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Minimum of x+1/x (Perimeter Square < Perim equal area rects)

  1. Apr 1, 2017 #1

    TheBlackAdder

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Gelfand - Algebra p.115 problem 264:
    Prove that a square has the minimum perimeter of all rectangles having the same area.
    Hint. Use the result of the preceding problem.

    2. Relevant equations
    Preceding problem: Prove that a square has the maximum area of all rectangles having the same perimeter.

    Perimeter of a square = Perimeter of a rectangle with an equal perimeter: $${4x} = 2(x-r)+2(x+r) = 2x-2r+2x+2r = 4x$$
    Area of that square: x²
    Area of that rectangle: (x-r)(x+r) = x²+rx-rx-r² = x²-r²

    Since r² is always positive, then x² > x²-r².

    3. The attempt at a solution
    My initial attempt, I did coincidentally try with the result of the preceding problem. Let's take the area of a square and let it be equal to the area of a rectangle: x² = x² - r²
    The perimeter of that square: 4x
    The perimeter of that rectangle: 2(x-r)+2(x+r) = 2x-2r+2x+2r = 4x
    So if the area of a square and a rectangle are equal, the perimeter of that rectangle is equal to the perimeter of the square. Then it dawned on me that x² = x² - r² is impossible.

    Second try:
    The area of a square = all rectangles with the same area: $$x² = (ax)(\frac{1}{a}x)$$
    The perimeter of that square: 4x
    The perimeter of those rectangles: $$2ax+\frac{2}{a}x$$
    $$2x(a+\frac{1}{a})$$

    Now I got to prove, for a positive a not equal to 1, that: $$a+\frac{1}{a} > 2 $$

    When that's proven, it shows that a square always has the minimum perimeter of all rectangles having the same area. (when a=1, it is the same square, not a rectangle)

    Here my first try was a proof by contradiction.
    $$a+\frac{1}{a} < 2$$
    $$\frac{1}{a} < 2 - a$$
    $$1 < 2a - a²$$
    $$0 < 2a - a² - 1$$
    Multiplied both sides by -1
    $$0 > a² - 2a + 1$$
    $$0 > (a-1)²$$
    Which is false because the square is always positive. So: $$4x < 2x(a+\frac{1}{a})$$
    Could someone lead me in the right direction towards finding other, perhaps easier proofs?
     
    Last edited: Apr 1, 2017
  2. jcsd
  3. Apr 1, 2017 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Use the arithmetic-geometric mean inequality. When are the two types of means equal?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Minimum of x+1/x (Perimeter Square < Perim equal area rects)
Loading...