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    Does anyone know how to prove (using calculus), that the quadrilateral with the largest area with a perimeter of 1000 metres is 250 x 250 metres?

    Similarly, is it possible to do likewise with an equilateral triangle of the same perimeter (i.e all of its sides are 333.33... metres).

    Finally, if anybody knows of any websites on calculus for 'beginners', it would be greatly appreciated.
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    A = x . y

    P = 2(x+y)
    1000 = 2(x+y)
    500 = x + y
    y = 500 - x

    A = x . (500 - x) = 500x - x²

    dA/dx = 500 - 2x = 0
    x = 250
    y = 500 - 250 = 250
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    Basically, you have to find everything in terms of one variable.

    So for a rectangle x by y, for example, area = xy 2x+2y=1000
    so 2y=1000-2x
    y=500-x
    area=x(500-x)=500x-x^2
    dA/dx=500-2x
    for a maximum, dA/dx = 0
    500-2x=0, x=250. 2x+2y=1000, y=250. Area =250*250 = 62500.
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    Similarly, is it possible to do likewise with an equilateral triangle of the same perimeter (i.e all of its sides are 333.33... metres).
    well an equilateral triangle of a constant perimeter has a constant area! it doesnt have a maximum or minimum area! as the angle is not a variable (60o) and as the area = ½.side.side.sin the angle between them = constant.
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    (Original post by yazan_l)
    A = x . y

    P = 2(x+y)
    1000 = 2(x+y)
    500 = x + y
    y = 500 - x

    A = x . (500 - x) = 500x - x²

    dA/dx = 500 - 2x = 0
    x = 250
    y = 500 - 250 = 250
    that assumes the quadrilateral is right angled...
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    Beaten, although yazan's (and my) working assumes that the quadrilateral must be rightangled.
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    Beaten twice
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    so... how does one show regular polygons have maximal area for their perimeter?
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    oh. you could use the AM GM inequality and keep shifting vertices...
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    (Original post by chewwy)
    that assumes the quadrilateral is right angled...
    hehe ... oops... my mistake!
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    Is there a simple formula for the area of a quad?
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    (Original post by chewwy)
    that assumes the quadrilateral is right angled...
    it is
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    I think you can prove with some fairly simple geometrical construction, superimposing a parallelogram on a rectangle of equal perimeter, that the area of the rectangle is greater; and no, I don't feel inclined actually to do it at the moment. I'm deeply into the claret. Rep to anyone who pops up a proof before I finish this bottle...
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    You can easily realise that such a quadrilateral will have angles less than 180, and you can see from symmetry that it should be some kind of parallelogram, and by further symmetry or by considering the area as a function of an angle, a rectangle is the shape you're after, and the calculus already mentioned shows it's a square.
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    as a total novice to calculus, I have managed to grasp some of what has been said here. However, if anyone knows any websites that explain it starting with the fundamentals, I would be grateful. It is all in aid of a GCSE Maths Coursework. Apparently, to get the top marks, you need to use caculus, and 'limits'?
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    http://www.karlscalculus.org/calculus.html

    Enjoy ...
 
 
 
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