calculus

An isosceles triangle is drawn with its vertex at the origin and its base parallel to the x-axis. The vertices of the base are on the curve 5y=25-x^2 Find the area of the largest such triangle.

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  1. your parabola is y = 5 - (1/5)x^2

    let the points of contact of the base of the triangle be (x,y) and (-x,y)
    Then the area is
    Area = xy/2
    = (x/2)(5-(1/5)x^2)
    = 5x/2 - (1/10)x^3

    d(Area)/dx = 5/2 - 3x^2/10
    = 0 for a max of Area
    3x^2/10 = 5/2
    x^2 = 25/3
    x = ± 5/√3 or appr. 2.88675

    sub back in area expression
    I had appr. 4.81125

    ( I tried x = 3 and x = 2.8 and they both gave me slightly smaller areas)

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  2. The base of the triangle should have been 2x, which makes the
    Area = xy
    = 5x - (1/5)x^3

    Can you figure out how that changes the answers?
    (hint: one stays the same, ...)

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  3. goodexcelent enteligent proud on u

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