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Which side of the triangle gives it maximu area?

  1. Mar 4, 2004 #1
    For my homework (actually its a bonus question) I was given an isocelese triangle with 2 sides lengths defined. Lets call it triangle ABC.

    Code (Text):

        A
        /\
       /  \
      /    \
     /      \
    /--------\
    B         C
     
    side AB is given 5 and side AC is also 5. I have to give BC a length for the triangle to have the maximum area.

    First I thought of squares. A square has more area then a rectangle with the same parameter. 5x5 is greater than 1x20 thus proves my point of squares. I thought that If the triangle would be equalateral it would have the greatest area. But I was wrong after expirementing with numbers.

    Then I thought since a square has the greatest area then half a square would be a triangle and therefore the right angle triangle should have the greatest area. Witch would have sqrt(50) as BC and an area of 12.5 according to my calculations (may be wrong).

    Any thoughts on what kind of triangle will have the greatest area?
     
    Last edited by a moderator: Mar 4, 2004
  2. jcsd
  3. Mar 4, 2004 #2

    Hurkyl

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    Can you write a formula for the area of the triangle in terms of the length of the base? That might suggest a way to find the maximum area...
     
  4. Mar 4, 2004 #3
    I don't know what you mean by that. the area of a triangle is base times hight divided by 2(bxh/2). You can use trig rations and sine/cos laws to find the hight of the triangle.
     
  5. Mar 4, 2004 #4

    Hurkyl

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    Well do that then; find the height of the triangle in terms of the length of the base.
     
  6. Mar 8, 2004 #5
    If you're lazy, you can use some trigonomery.
    S=(1/2)bcSinA
    Which angle has the biggest sine? pi/2 (+k2pi).
     
  7. Mar 8, 2004 #6

    ShawnD

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    Just use some simple calculus (or cheat and just graph it on your calculator).

    Look at this picture here
    triangle

    Now based on that diagram I can make 2 formulas:

    [tex]A = xy[/tex] since x is only half the base here

    [tex]y = (5^2 - x^2)^\frac{1}{2}[/tex]

    Combine them to get this

    [tex]A = x(5^2 - x^2)^\frac{1}{2}[/tex]

    Now differentiate it

    [tex]\frac{dA}{dy} = (5^2 - x^2)^\frac{1}{2} - \frac{y^2}{(5^2 - x^2)^\frac{1}{2}}[/tex]

    Make that formula equal to 0 and you get x.
    Since my diagram had x as only HALF of the base length, double what you get for x.

    I get the base as being 7.071 long. That gives the triangle and area of 12.5


    Now about what kishtik said
    That gives the exact same area. (1/2)(5)(5) = 12.5

    Give the teacher both answer and you might get extra bonus marks
     
    Last edited: Mar 8, 2004
  8. Mar 8, 2004 #7
    This isn't that hard. Flip the triangle so that a given side is on the bottom. (Base) You know that the formula for area is base x height x 1/2.

    Now pivot the other given side until you get the most height out of it. The base is given, you have the most height you can possibly get in ONE position only. Is this sinking in?
     
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