# Math Help - Maxima och trig equation

1. ## Maxima och trig equation

Hi

The goal is to find the angle x that gives the largest area.
I have an area expression:

$A(x)= 2L^{2}\cdot sin(\frac{x}{2})+L^{2}\cdot cos(\frac{x}{2}) \cdot sin(\frac{x}{2})$ where L is a constant.

If the derivative below is wrong, let me know. I did use some trig identities.

$A'(x)=L^{2}(cos(\frac{x}{2})+\frac{1}{2}\cdot cos(x))$

So basically, how do I find the x the gives a maxima?

Thanks

2. Hello, Twig!

The goal is to find the angle $x$ that gives the largest area.

I have an area expression:

$A(x)\:=\: 2L^{2}\sin(\tfrac{x}{2})+L^{2}\cos(\tfrac{x}{2})\s in(\tfrac{x}{2})$ .where $L$ is a constant.

If the derivative below is wrong, let me know. I did use some trig identities.

$A'(x)\:=\:L^2\left(\cos(\tfrac{x}{2})+\tfrac{1}{2} \cos(x)\right)$ .
This is correct, but . . .

So basically, how do I find the $x$ the gives a maxima?

Our equation was: . $A'(x) \;=\;L^2\left(\cos\tfrac{x}{2} + \tfrac{1}{2}\cos^2\!\tfrac{x}{2} - \tfrac{1}{2}\sin^2\!\tfrac{x}{2}\right) \;=\;0$

Multiply by $\frac{2}{L^2}\!:\quad 2\cos\tfrac{x}{2} + \cos^2\!\tfrac{x}{2} - \sin^2\!\tfrac{x}{2} \;=\;0$

Replace $\sin^2\!\tfrac{x}{2}$ with $1 - \cos^2\!\tfrac{x}{2}\!:\quad 2\cos\tfrac{x}{2} + \cos^2\!\tfrac{x}{2} - (1 - \cos^2\!\tfrac{x}{2}) \;=\;0$

. . And we have: . $2\cos^2\!\tfrac{x}{2} + 2\cos\tfrac{x}{2} - 1 \;=\;0$ . . . a quadratic in $\cos\frac{x}{2}$

Quadratic Formula: . $\cos\tfrac{x}{2} \;=\;\frac{-2 \pm\sqrt{12}}{4} \;=\;\frac{-1\pm\sqrt{3}}{2}$

We find that: . $\cos\frac{x}{2} \:=\:\frac{-1 - \sqrt{3}}{2} \:=\:-1.36\hdots$ . has no real solutions.

And that: . $\cos\frac{x}{2} \:=\:\frac{-1+\sqrt{3}}{2} \quad\Rightarrow\quad \frac{x}{2} \:=\:68.529857^o \quad\Rightarrow\quad x \:=\:137.0585971^o$

. . Therefore: . $x \;\approx\;137^o$

3. Hi!

Thank you so much!

This approach was much better.

I tried to re-write with half-angle formulas and stuff, I thought it should work but it didn´t. =)