# Problem of the week helpp

1. Dec 14, 2004

### calvinnn

I posted this before, but i messed up on the wording, so heres my repost. My teacher gave us our problem of the week (which was previously on our test), and i have no idea on how to solve it. All i know is that i will probably need to differentiate something, and then find the criticle numbers. OK, so here it goes:

Use Calculus to prove which vertex angle gives an isosceles triangle the greatest area

Figure Below
THank You

#### Attached Files:

• ###### Figure Here.JPG
File size:
4.2 KB
Views:
66
2. Dec 14, 2004

### learningphysics

3. Dec 14, 2004

### Tide

I think you were already given the solution someplace else but perhaps you're looking for more detail?

The first question is do you have any contraints in the problem? For example, is the perimeter of the triangle fixed? I'd guess that the length of the equal sides is fixed since they are labled "K" which suggests a constant.

In that case the area of the triangle is

$$A = K^2 \sin \frac {\theta}{2} \cos \frac {\theta}{2} = \frac {1}{2}K^2 \sin \theta$$

where $\theta$ is the vertex angle. Does that help?

4. Dec 14, 2004

### calvinnn

thanks

things seem a bit clearer
Thanks