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Maximizing area of a triangle

  1. Oct 25, 2006 #1
    So here is the problem. I did it but just wanted to make sure it is correct... does the solution look ok to you guys.

    problem: an Isosceles triangle has its vertex at the origin and its base parallel to the x-axis with the vertices above the x-axis on the curve y = 27-x^2. find the largest area the triangle can have.

    solution:So I assumed the base of the triangle will be the points (x, y) and (-x, y).

    Therefore height of the triangle will be y
    length of the base will be x+x = 2x

    Area = 1/2 * base * height
    = 1/2 * 2x * y

    now we know that y = 27 - x^2
    so this area eqn becomes
    A = x(27 - x^2)
    A = 27x - x^3
    A' = 27 - 3x^2 = 0
    so the critical points are x = 3 and x = -3

    using the sign chart i figured out that i have to use x = 3 because that is the local maximum.

    and when I plug x = 3 back into A = x(27 - x^2) we get A = 54 unit^2

    Can some one verify that my solution is correct? I would really appreciate it.
  2. jcsd
  3. Oct 25, 2006 #2

    Andrew Mason

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    Correct analysis and correct result.

  4. Oct 25, 2006 #3


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    One small point is that you should be more careful about what range you're trying to find a maximum over. For example, at x=-10, A=+730, so why isn't this a better solution? The reason is that you're looking only for solutions with 0<=x<=r, where r is the smallest positive root of 27-x^2. You should compare all local maxima in this range along with the values at the endpoints and pick the largest of these values, which will give what you found.
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