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Math Help - the biggest area

  1. #1
    Super Member dhiab's Avatar
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    the biggest area

    What is the biggest area of a triangle can be drawn inside the ellipse equation :  \left( {\frac{x}{3}} \right)^2 + \left( {\frac{y}{7}} \right)^2 = 1<br /> <br />
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    Quote Originally Posted by dhiab View Post
    What is the biggest area of a triangle can be drawn inside the ellipse equation :  \left( {\frac{x}{3}} \right)^2 + \left( {\frac{y}{7}} \right)^2 = 1
    For the circle x^2+y^2=a^2, the largest triangle that can be drawn inside it is the equilateral triangle with side a\sqrt3, whose area is \tfrac{3\sqrt3}4a^2.

    The ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 is obtained from the circle by the linear transformation with matrix \begin{bmatrix}1&0\\0&b/a\end{bmatrix}. (In non-technical language, you get the ellipse from the circle by leaving the x-coordinate unchanged and expanding or shrinking the y-coordinate by a factor b/a.) This transformation alters areas by a factor b/a. So the largest triangle that can be inscribed in the ellipse has area \tfrac{3\sqrt3}4ab, or 63\sqrt3/4 square units when a=3 and b=7.
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    If I may, here is another way using polar coordinates.

    The area of the triangle can be expressed as

    A=\frac{1}{2}(2acos{\theta})(b+bsin{\theta})

    That is just \frac{1}{2}\text{base}\cdot \text{height}

    Differentiate, set to 0 and solve for theta gives us \theta=\frac{\pi}{6}

    Sub this into A and we get A=\frac{3\sqrt{3}ab}{4}
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    Quote Originally Posted by galactus View Post
    If I may, here is another way using polar coordinates.

    The area of the triangle can be expressed as

    A=\frac{1}{2}(2acos{\theta})(b+bsin{\theta})

    That is just \frac{1}{2}\text{base}\cdot \text{height}

    Differentiate, set to 0 and solve for theta gives us \theta=\frac{\pi}{6}

    Sub this into A and we get A=\frac{3\sqrt{3}ab}{4}
    That method works fine provided that you assume that the base of the triangle is horizontal. What I found interesting (and unexpected) about this problem is that the same maximum area 3\sqrt{3}ab/4 can be achieved with a triangle whose base has any given orientation. That follows easily if you think of the ellipse as a "squashed circle", but I think it would be hard to prove by any other method.
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  5. #5
    Super Member dhiab's Avatar
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    Quote Originally Posted by Opalg View Post
    That method works fine provided that you assume that the base of the triangle is horizontal. What I found interesting (and unexpected) about this problem is that the same maximum area 3\sqrt{3}ab/4 can be achieved with a triangle whose base has any given orientation. That follows easily if you think of the ellipse as a "squashed circle", but I think it would be hard to prove by any other method.
    Hello Thank you ARE YOU THE DETAILS FOR FORMULA AERA
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