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max area of isosceles triangle

  • 09-05-2006 12:43PM
    #1
    Elfman
    Registered Users, Registered Users 2 Posts: 221 ✭✭


    Im looking for a formmul;ar i could use in a computer program to find the max area of an isosceles triangle where the sides adsd up to a given length.

    Hope im describing this right all help would be great thanks you


    -Elfman


Comments

  • #2
    breadmonkey
    Registered Users, Registered Users 2 Posts: 3,608 ✭✭✭breadmonkey


    Give what you are trying to maximize a symbol A or something.

    Express A as a function of the side lengths a and b, using the standard formula for the area of a triangle.

    Take 'a' as the length of the equal sides and 'b' as the other side. The perpendicular height = sqrt(a^2 - b^2/4) by pythagoras.

    Therefore A = (b/2)*sqrt((a^2 -b^2)/4)) -->(equation 1)

    You know the perimeter P so you can express a as a function of b and vice versa. P = a + a + b

    So b = P - 2*a

    Substitute this back into the formula for the Area and you now have A as a function of a single variable, a.

    A = ((P - 2*a)/2)*sqrt(4*a^2 -4*a*P -P^2)

    Differentiate with respect to a, equate to 0 and you should get that for maximum area,

    EDIT: Made a mistake doing this the first time and amn't arsed doingg it again!

    Then ram this back through the original formula for A, ie. equation 1, and that's the answer.

    I think. I hope I haven't made a balls of that.


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