What is the area of the largest rectangle that can be inscribed in an isosceles triangle with side lengths $8$, $\sqrt{80}$, and $\sqrt{80}$? assume that one side of the rectangle lies along the base of the triangle.

First let's solve the general case for an isosceles triangle of base 2 and height k. If the triangle is drawn so the base is on the x-axis and the apex is on the y-axis, the equation for the line containing the right side is
  y = -k(x-1)
Then a rectangle with x-dimension w will have an area that is the product of this width and the height y = -k((w/2)-1).
  area = w(-k(w/2 -1)) = (-k/2)w² +kw
The derivative of area with respect to w will be zero where the area is a maximum:
  d(area)/dw = 0 = -kw +k
  w = 1 . . . . . . add kw and divide by k
Using the formula for area, we find
  area = 1(-k(1/2-1)) = k/2
This value is 1/2 the total area of the triangle we started with.

If the side lengths of your triangle are 8, √80, and √80, the height will be given by the Pythagorean theorem as
  h = √((√80)² - (1/2·8)²) = √(80-16) = 8
Your triangle's area will be
  triangle area = (1/2)(8)(8) = 32

The maximum area of the inscribed rectangle will be half this value,
  16 square units