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Show that of all the rectangles with a given area, the one with smallest perimeter

  1. Aug 3, 2008 #1
    1. The problem statement, all variables and given/known data
    (11)
    a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square.
    b) Show that of all the rectangles with a given perimeter, the one with greatest area is a square.


    2. Relevant equations
    - Differentiation rules
    - Superhuman powers


    3. The attempt at a solution
    I'm completely lost on this one. I honestly don't even know where to start. Of all the rectangles with a given area? What?

    So let's introduce some variables.

    Say we have 3 rectangles. Their areas are a1, a2, a3. Then their perimeters are p1, p2, p3. Now if p2 > p1 <p3, i.e. p1 is the smallest of the three perimeters, then the rectangle 1, with area a1 and perimeter p1 is square. I get the question.. but I still don't have a clue about where or how to start.

    Since b) is very similar, the same goes for b).
     
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  3. Aug 3, 2008 #2

    turbo

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    Re: Show that of all the rectangles with a given area, the one with smallest perimete

    Can you break the question down for yourself by using examples? You can have very skinny rectangles with lots of perimeter and little area, for example. This problem also extends to other 2-d shapes. What is the largest area that can be encompassed by a 2-d shape? How is this shape approximated in regular 2-d shapes?
     
  4. Aug 3, 2008 #3

    tiny-tim

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    Hi illjazz! :smile:

    Let's do it step by step …

    If a rectangle has sides x and y:

    i] what is the area, A?

    ii] what is the perimeter, P?

    Now if A is given (fixed), what is the smallest value of P (in terms of x and y)? :smile:
     
  5. Aug 3, 2008 #4
    Re: Show that of all the rectangles with a given area, the one with smallest perimete

    The largest area that can be encompassed by a 2d shape? Um.. I'd say infinity. A shape is approximated by taking a 2d object with n "sides" and then increasing the number of sides long enough to reach the desired level of precision, I guess?

    Ok. So we have the sides, x and y. Then the area A is

    i] A = xy

    and the perimeter is

    ii] P = 2x + 2y

    I guess the smallest value of P then would be either 4x or 4y. Huh! This makes sense :). But I just "described" it.. I didn't make any equations :/
     
    Last edited: Aug 3, 2008
  6. Aug 3, 2008 #5

    Defennder

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    Re: Show that of all the rectangles with a given area, the one with smallest perimete

    Well then next you should express P in terms of either x or y only, then optimise the area using calculus. Then you'll get the answer.
     
  7. Aug 4, 2008 #6

    tiny-tim

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    Hi illjazz! :smile:
    erm … no, it doesn't … it doesn't tell you anything at all!

    The problem is:

    Given xy = A (a constant), for what value of x (in terms of A) is 2x + 2y smallest? :smile:
     
  8. Aug 4, 2008 #7

    HallsofIvy

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    Re: Show that of all the rectangles with a given area, the one with smallest perimete

    Yes, that is the question. Also, since you know that xy= A, y= A/x where A is some constant. That lets you replace the "y" in P= 2x+ 2y so that the perimeter, P(x), is a function of x only. Now, the crucial question: do you know how to use calculus to find where a function takes on a maximum or minimum value?
     
  9. Aug 4, 2008 #8

    tiny-tim

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    Hi HallsofIvy! :smile:

    Or, if you don't like calculus :frown:, you can complete the square.

    (Hint: try a difference of two square-roots :wink:)
     
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