[SOLVED] An Applied Trig/Max and Min Problem

Hi there,

I'm currently struggling with the following question from my textbook.

"Points A and B lie on a circle, centre O, radius 5 cm. Find the value of angle B that produces a maximum area for triangle AOB."

I know that I need to find the area and then the derivative and set it to zero but I'm not sure how to find the base and height of the triangle to get the area.

Thanks!

Re: An Applied Trig/Max and Min Problem

Use the formula for the area A of a triangle:

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

You don't even need calculus to find the angle that maximizes A...

Re: An Applied Trig/Max and Min Problem

Okay, right. I was just over thinking it.

So then, it's just $\displaystyle \pi/2$?

Thanks again MarkFL

Re: An Applied Trig/Max and Min Problem

Re: An Applied Trig/Max and Min Problem

Quote:

Originally Posted by

**MarkFL2** you got it! ;)

Except that cannot form a triangle with both A and B on the circle and andgle OBA=90 degrees!.

Draw a picture!

CB

Re: An Applied Trig/Max and Min Problem

Quote:

Originally Posted by

**MarkFL2** Use the formula for the area A of a triangle:

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

You don't even need calculus to find the angle that maximizes A...

What are a and b, how are they related to the angle OBA?

Also do not use the same symbol for two different thing in the same problem: A - area of and label of a vertex of the triangle.

CB

Re: An Applied Trig/Max and Min Problem

My apologies for my sloppiness.

Using coordinate geometry:

Putting point O at (0,0), point A at (5,0) and B at $\displaystyle \left(5\cos\theta,5\sin\theta\right)$ where $\displaystyle \theta=\angle AOB$ then the area T of the triangle is:

$\displaystyle T=\frac{1}{2}(OA)(OB)\sin(\theta)$

Re: An Applied Trig/Max and Min Problem

Quote:

Originally Posted by

**MarkFL2** My apologies for my sloppiness.

Using coordinate geometry:

Putting point O at (0,0), point A at (5,0) and B at $\displaystyle \left(5\cos\theta,5\sin\theta\right)$ where $\displaystyle \theta=\angle AOB$ then the area T of the triangle is:

$\displaystyle T=\frac{1}{2}(OA)(OB)\sin(\theta)$

OK the area is maximised when your $\displaystyle \theta=\pi/2$ , but that is not what was asked for, what was asked for is angle OBA (which is now trivial to find, but your original presentation cannot have been anything but confusing to the OP).

CB