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Math Help - Global Maximum for area of a triangle.

  1. #1
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    Global Maximum for area of a triangle.

    Hello,

    I need to find the length of the base of a (isoceles) triangle (x) to get the maximum area in terms of the fixed sides (L)

    so far I have found the critical points of the derivative

    Area = (1/2) xh

    Where h = ( L^2 - (x^2)/4)^1/2

    L = Fixed lengths of triangle = constant

    So when derivative = 0, we have
    Area' = x = L* (2)^1/2

    I just need to know how i can test if this is a global maximum of the fucntion.
    do i just find the limits when x goes to 0 and infinity? What does it mean when i do?

    Cheers
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  2. #2
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    i think taking the second derivative allows you to ascertain whether it is a global min or max, like, if the second derivative is positive, i think its minimum and if the second deriv is negative, i think it will be a global max
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  3. #3
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    No, that only tests for local max and min. To find Global max and min just calculate the area at each critical point and at the "endpoints", the base being 0 or \infty (those are easy- the area is 0 so clearly not a maximum). The largest of those is the global maximum.

    But I have no idea what you mean by "Area'= x= L*(2)^1/2" At the maximum area, Area'= 0, of course. Are you saying the maximum area occurs when x= L*(2)^1/2? If so, that is not what I got.
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  4. #4
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    I'm getting the same answer as the original poster, i.e., x=L \sqrt{2}.

    By the way, the the bounds of the possible values for x would be 0<x<2L. You may recall that the sum of the lengths of two sides of a triangle must be greater than the length of the third side.

    In any case, you've determined that there is only one value of x where the derivative equals zero, and that the derivative is continuous, so we know that if the function has a global maximum, it will occur at x=L\sqrt{2}.

    You have several ways to show that a local maximum occurs (which also proves that a global maximum occurs because there is only one critical point and the function is continuous). The method of finding the limit as x \to 0^+ and x \to 2L^- will work. You just need to demonstrate that the areas at these endpoints are less than the area you get at x=L \sqrt{2}. Like HallsofIvy said, though, it's zero at the endpoints, so it's clear that the solution found is in fact a maximum.
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