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alg 2

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a rectangle is to be inscribed in a isosceles triangle of height 8 and base 10. Find the greatest area of such rectangle.

  • alg 2 - ,

    Did you make a sketch?

    Draw triangle ABC so that AB = AC
    Draw the height AD, D on BC
    Let P be the point where the rectangle touches AC
    Draw PE perpendicular to BC, E on BC

    let DE = x, then EC = 5-x
    let PE = y
    by similar triangles:
    y/(5-x) = 8/5
    y = (40-8x)/5

    Area of rectangle = 2xy
    = 2x(40-8x)/5
    = (80x - 16x^2)/5

    d(Area)/dx = (80 - 32x)/5 = 0 for max of area
    32x = 80
    x = 80/32 = 5/2
    y = (40 - 8(2.5))/5 = 4

    greatest area = 2xy = 2(5/2)(4) = 20

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