Calculus

SOS...help!!! please email me at 969e221 at g mail. com here is the question:
Find the trapezoid of largest area that can be inscribed in the upper half of the ellipse x2/9 + y2/4 = 1 where the lower base of the trapezoid is on the x-axis and is 6 units long. I need to graph the ellipse and draw an inscribed trapezoid. Write and area formula for the trapezoid in x. Differentiate, find critical numbers, and apply an appropriate derivative test to show that you have found the trapezoid of largest area. HELP!!!!!!!!!!!!!!!!

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asked by Buster
  1. well shucks. The base of the trapezoid is just the long axis of the ellipse, so you want to find two points (-x,y) and (x,y) on the ellipse which gives maximum area

    a = 1/2(2x+6)y = (x+3)y

    Now, we know that x^2/9 + y^2/4 = 1, so
    y^2 = 4(1 - x^2/9)
    so y = 2/3 √(9-x^2)

    a = 2/3 (x+3)√(9-x^2)
    da/dx = -(2x^2+3x-9)/√(9-x^2)
    = -(x+3)(2x-3)/√(9-x^2)
    This achieves a max at x=3/2.
    We don't really need to find that a" < 0 at x=3/2, since with x = -3, a=0, a min.

    Anyway, now just solve for y at x = 3/2 to find the area. The trapezoid has a top base of 3 and a bottom base of 6, with height = y.

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    posted by Steve
  2. omg thanks

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    posted by Buster

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