# Math Help - the biggest area

1. ## 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

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2. Originally Posted by dhiab
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.

3. 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 $A=\frac{3\sqrt{3}ab}{4}$

4. Originally Posted by galactus
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.

5. Originally Posted by Opalg
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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