Question 1012602
area of triangle EAF = x^2 / 2
area of triangle EBC = ((1-x)*1) / 2 = (1-x) / 2
area of square ABCD = 1 * 1 = 1
area of quadrilateral CDEF(y) = area of square ABCD - area of triangle EAF - area of triangle EBC
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y = 1 - x^2/2 - (1-x) / 2
y = (2 - x^2 - (1-x)) / 2
y = (1 +x -x^2) / 2
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this is a parabola that opens downward, so we want the value of x at the parabola's peak.
x = -b / 2a = (-1/2) / (2 * (-1/2)) = 1/2
now substitute 1/2 for x in the equation for y
y = (1 +(1/2) - (1/2)^2) / 2 = (5/4) / 2 = 5/8
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the greatest possible area for quadrilateral CDEF is 5/8
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here is the graph of the area of quadrilateral CDEF
{{{graph(300, 200, -3, 3, -3, 3, (1 +x -x^2) / 2)}}}