What is the largest possible area for a right triangle with hypotenuse of 15cm, and what are its dimensions? (using calculus)?

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  • 4 weeks ago

    let a and b be the other two sides

    area = (1/2)ab

    a² + b² = 15²

    a = √(225 – b²)

    area = (1/2)b√(225 – b²)

    take derivative, set equal to 0 and solve for b

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  • david
    Lv 7
    4 weeks ago

    legs are x and y

      x^2 + y^2 = 15^2

        y = sqrt[225 - x^2]

     A = (1/2)xy 

    A(x) = (x/2)*sqrt[225 - x^2]

    A'  =  (x/2)(1/2)(-2x)/sqrt[225 - x^2]  +  (1/2)sqrt[225 - x^2] = 0

        (-x^2)/2 + (1/2)(225 - x^2) = 0

       x^2 + x^2 = 225

        x^2 = 225/2

        x = 15/sqrt2  <<<  do a little algebra and y = same = 15/sqrt2

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  • 4 weeks ago

    Intuitively the answer is when you have a 45-45-90 triangle where the legs are equal and have a value of 15/√2.

    But if you want to be a little more rigorous you can resort to calculus.

    Let a be the measure of one leg.

    Let b be the measure of the other leg.

    By the Pythagorean Theorem:

    a² + b² = 15²

    a² + b² = 225

    a² = 225 - b²

    a = √(225 - b²)

    The formula for the area of a right triangle is half the product of the legs.

    A = ½ab

    Substitute in for a and you'll have a function for the area, based on the value of b.

    f(b) = ½√(225 - b²)b

    Take the derivative:

    f'(b) = ½(225 - 2b²)/√(225 - b²)

    Set that to zero. We only really care about the numerator:

    225 - 2b² = 0

    2b² = 225

    b² = 15² / 2

    b = √(15²/2)

    b = 15/√2

    And you can then find out that a = 15/√2

    Answer.

    Both legs are 15/√2 

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  • 4 weeks ago

    The largest area is when the legs are each 15 / sqrt(2).

    • david
      Lv 7
      4 weeks agoReport

      true ... but how do you use calculus to prove it?

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