# maximum area

• Dec 5th 2010, 05:06 PM
purplelollipop27
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.Attachment 19979
• Dec 5th 2010, 05:18 PM
skeeter
Quote:

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.Attachment 19979

for any isosceles triangle with equal sides L and vertex angle $\theta$ ...

$A = \frac{1}{2} L^2 \sin{\theta}$
• Dec 5th 2010, 05:21 PM
dwsmith
$\displaystyle A=\frac{1}{2}bh$

Consider 3 possibilities:

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

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

Theta =90
Theta < 90
Theta > 90
• Dec 5th 2010, 06:16 PM
SammyS
Maximize area of isosceles triangle.
Quote:

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.Attachment 19979
Modify your attached figure so that the equal sides have length, $\ell$, rather than 1. Then the altitude, $\displaystyle h$, is given by $\displaystyle h= \ell\cos{\theta\over2}$ and the base, $\displaystyle b$, is given by $\displaystyle b=2\ell\sin{\theta\over2}$
The area is: $\displaystyle A={1\over2}bh={1\over2}\left(2\ell\sin{\theta\over 2}\right)\left(\ell\cos{\theta\over2}\right)$
Use $\displaystyle \sin(2x)=2\sin(x)\cos(x)$, with $\displaystyle x={\theta\over2}$ to simplify this to
$\displaystyle A={1\over2}\,\ell\, ^2\sin\theta$.
Maximize the area with respect to $\displaystyle\theta$.