Calculus 1: Max-Min Problems (22 of 30) Maximum Rectangle Inside a Semi-Circle

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  • Опубликовано: 18 ноя 2024

Комментарии • 15

  • @adegliangeli
    @adegliangeli 7 лет назад +3

    It's easier if you take the angle alpha. Then you have A=2R^2cos(alpha)sin(alpha). Amax is reached when you have alpha = pi/4.

  • @kumaravelannamalai979
    @kumaravelannamalai979 4 года назад +1

    Easy to understand

  • @katyperryaf8876
    @katyperryaf8876 6 лет назад +4

    Great video as always, I did have a small concern though, when taking the d/dx of sqrt(r^2-a^2) why was the d/dx of r^2 within g(x) not at all acknowledged, If I understand the chain rule correctly when taking the d/dx g(x) shouldn't it have been =(2r-2a)as appose to (-2a). I understand I'm wrong in assuming this but I'd like to know why, any clarification would be much appreciated

    • @MichelvanBiezen
      @MichelvanBiezen  6 лет назад +4

      In this problem, R is a constant and "a" is a variable.

    • @katyperryaf8876
      @katyperryaf8876 6 лет назад +3

      Clears up my confusion completely Thank you.

  • @xavier3153
    @xavier3153 4 года назад +1

    is that quadrilateral which is inscribed a square

  • @scottmenace1
    @scottmenace1 5 лет назад +3

    not all hero wear cape

  • @dukenukem2578
    @dukenukem2578 5 лет назад +3

    all that just for r^2. wow thanks obama

  • @ull893
    @ull893 7 лет назад +1

    Amazing!!

  • @piyushborse
    @piyushborse 11 месяцев назад +1

    and this question cam in cat 2023 slot 3😮‍💨😞

    • @MichelvanBiezen
      @MichelvanBiezen  11 месяцев назад +1

      We are not familiar with the CAT test.

  • @immamachine2436
    @immamachine2436 3 года назад +1

    amazin

  • @matthewmonceda6582
    @matthewmonceda6582 5 лет назад

    i dont get it

  • @aathithyan_k_k8600
    @aathithyan_k_k8600 7 лет назад +4

    sorry that is not triangle that is rectangle