Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    Correct analysis and correct result.

  4. Oct 25, 2006 #3


    User Avatar
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook