![]() Using differential calculus, find the absolute maximum (or minimum). Do the Second derivative test to make sure it's a maximum: A''(x) We’ll need this in several places so we can’t forget this. ![]() Since x is a length, so it's a positive value, so x = 1/2. g (c) f (c) 2f(c) By assumption we know that f(c) exists and f(c) > 0 and therefore the denominator of this will always exist and will never be zero.After solve the first derivative by applying product rule, we get: x=-1 or 1/2. Set A'(x) = 0 to find the critical point first.Back to the beginning, since we are to find the largest area, so it’s saying we are to find the Maximum value of the function A(x).So the final function looks like this: A(x) = 1/2.Since (x, y) is a point of the circle, so the circle's Standard formulashould work: x² + y² = r².So the y has to transform to the term of x. We need to form a function as Area in term of x, means the Area would change with the change of x.The h is hight of the trapezoid, which is equal to the top vertex's y value.The length of Top Base should be b₂ = x - (-x) = 2x.The other vertex on the Top base should be (-x, y).Since the two bases are parallel, so the Top Base is symmetric as well, means it has equal distance to both sides’ vertices, which could conclude that:.Assume one of the vertex on the Top base is (x, y).symmetric about the Y-axis (because two bases are parallel).equal to the Diameter (because it’s the longest base’s length).b₁ is the bottom base, which is equal to 2r = 2, because the largest-area-trapezoid inscribed in circle MUST:.First form the equation of trapezoid’s area: A = 1/2.Get some clue from the result first, then see what’s missing and find a way to get it.With knowing all these conditions above, you could start to abstract the information into equations.And it means it's absolute symmetric about the X-axis. Since it’s a trapezoid, and inscribed in a circle, then IT HAS TO BE A ISOSCELES TRAPEZOID.Which means the length of bottom base should be twice radius: b₁ = 2 A trapezoid of maximum area inscribed in the semicircle will have its base on the X-axis.Know that, a quadrilateral CAN be inscribed in a circle or even a semicircle, which means 4 vertices are all on the circle.Remember that a trapezoid has to have TWO BASES to be parallel.For this Trapezoid inscribed in circle problem, you really want to draw it out before anything else.Refer to this animated tool from Geogebra that I created for this problem.
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