You have a 4inch by 6inch notecard and you intend to make a triangle out of the notecard by using one long side (call it AB) as the base of your triangle and by then placing the third vertex (point E) somewhere on the opposite side (call it CD).

Where should you place point E so that the resulting triangle has a minimum perimeter?
Where should you place point E so that the resulting triangle has a maximum perimeter?

If you believe the perimeter is the same regardless of E's position, then prove this statement.

If you believe the perimeter changes as E's position changes, determine a function for the perimeter of the resulting triangle given E's position on the segment CD. Make sure to clearly define your variables and properly explain your resulting function.


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  1. Hahaha! Is this for Educ 439?
    Either way the answer is:
    Perimeter F(x)= 6+(Sqrt((X^2)+16))+(Sqrt((X-6)^2)+16))

    The goal of the function is finding the hypotenuse of the two triangles not being used by the triangle you created. The last side, the base, will always be 6. Obviously, X can only be between 0-6.
    Maximum = F(0),F(6)= 17.211 (approx)
    Minimum = F(3) = 16

    Hope this helps!

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