ms sue

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triangle base is (4-x)cm and its height is (2x+3)cm.what will be the maximium area?

  • ms sue - I don't know -

    Sorry.

  • ms sue -

    Area = (1/2)base x heigh

    = (1/2)(4-x)(2x+3)
    = (1/2)(12 + 5x - 2x^2)

    d(Area)/dx = (1/2)(5 - 4x)
    = 0 for a max area

    4x = 5
    x = 5/4

    using x = 5/4 = 1/25
    Maximum Area = (1/2)(4-5/4)(5/2+3) = 7.5625
    or appr 7.5625

    testing: our answer t= 5/4 = 1.25
    let x = 1.24 , area = (1/2)(4-1.24)(2.48+3) = 7.5624 , a bit smaller
    let x = 1.26, area = (1/2)(4-1.26)(2.52+3) = 7.5624 , a bit smaller again

    all looks good at max area = 7.5625

  • ms sue -

    can you explain how you got 1/2 x 5-4x ?

  • ms sue -

    Since this is a typical Calculus question, I assume you knew Calculus.
    I took the derivative.

  • ms sue -

    You could start with my area equation

    area = (1/2)(12 + 5x - 2x^2)
    = (1/2)(-2)(x^2 - 5/2x -6)
    = -(x^2 - (5/2)x + 25/16 - 25/16 - 6) ----I competed the square
    = -( (x - 5/4)^2 - 25/16 - 6)
    = (x - 5/4)^2 + 25/16 + 6
    = -(x-5/4)^2 + 121/16

    area = -(x-5/4)^2 + 121/16

    max area = 121/16 or 7.5625
    when x = 5/4

    just as in my first solution.

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