1. ## maximum area

An isoceles triangle has two sides of length (l), and the angle between the sides has measure theta. Determine the angle theta that maximizes the area of the triangle, and find the maximum possible area.

2. Originally Posted by purplelollipop27
An isoceles triangle has two sides of length (l), and the angle between the sides has measure theta. Determine the angle theta that maximizes the area of the triangle, and find the maximum possible area.
for any isosceles triangle with equal sides L and vertex angle $\displaystyle \theta$ ...

$\displaystyle A = \frac{1}{2} L^2 \sin{\theta}$

3. $\displaystyle \displaystyle A=\frac{1}{2}bh$

Consider 3 possibilities:

$\displaystyle \displaystyle \frac{\theta}{2}+\frac{\theta}{2}=\theta$

That assume the angle isn't split at first to solve the whole area of the triangle.

Theta =90
Theta < 90
Theta > 90

4. ## Maximize area of isosceles triangle.

Originally Posted by purplelollipop27
An isosceles triangle has two sides of length (l), and the angle between the sides has measure theta. Determine the angle theta that maximizes the area of the triangle, and find the maximum possible area.
Modify your attached figure so that the equal sides have length, $\displaystyle \ell$, rather than 1. Then the altitude, $\displaystyle \displaystyle h$, is given by $\displaystyle \displaystyle h= \ell\cos{\theta\over2}$ and the base, $\displaystyle \displaystyle b$, is given by $\displaystyle \displaystyle b=2\ell\sin{\theta\over2}$
The area is: $\displaystyle \displaystyle A={1\over2}bh={1\over2}\left(2\ell\sin{\theta\over 2}\right)\left(\ell\cos{\theta\over2}\right)$
Use $\displaystyle \displaystyle \sin(2x)=2\sin(x)\cos(x)$, with $\displaystyle \displaystyle x={\theta\over2}$ to simplify this to
$\displaystyle \displaystyle A={1\over2}\,\ell\, ^2\sin\theta$.
Maximize the area with respect to $\displaystyle \displaystyle\theta$.