Maximize area depending on angle

Question

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How can I choose the angle $\alpha$ so that the total area of the figure is maximized?

Let's call the line length for $a$ and the base of the triangle with its top angle $\alpha$ for $2b$. The height of this triangle we call $c$.

Then solve $b$ in $\alpha$: $\sin \left( \frac{\alpha}{2} \right) = \frac{b}{a} \Leftrightarrow b = a\sin\left(\frac{\alpha}{2}\right)$

And solve $c$ in $\alpha$ $\cos \left( \frac{\alpha}{2} \right) = \frac{c}{a} \Leftrightarrow c = a\cos\left(\frac{\alpha}{2}\right)$

The area is: $a\times b+ \frac{2bc}{2}\times 2 = 2ab+bc$

Insert $b$ and $c$ gives:

$2a \left(a\sin\left(\frac{\alpha}{2}\right)\right) + \left(a\sin\left(\frac{\alpha}{2}\right)\right)\left(a\cos\left(\frac{\alpha}{2}\right)\right) $

Since all $a$ is equal set $a = 1$: $2\left(\sin\left(\frac{\alpha}{2}\right)\right) + \left(\sin\left(\frac{\alpha}{2}\right)\right)\left(\cos\left(\frac{\alpha}{2}\right)\right) $

Now if I derivate this I get $2cos\left(\frac{\alpha}{2}\right)$ And that's clearly wrong, so I don't need to continue my calculations.... Where did I screw up?

I exchanged your "figure" with a better one. Please check if it's correct! — Daniel R, Oct 21 '13 at 11:29
Thanks, it is correct! Note that $\alpha$ is the angle $a$ to $a$ and not $a$ to $c$. — iveqy, Oct 21 '13 at 11:54

Daniel R · Answer 1 · 2013-10-21T12:12:11.150

1

Your expression for the area is correct, but the differentiation seems to have gone wrong. Here's one way to do it:

$$A = a^2\left(2\sin \frac{\alpha}2 + \sin \frac{\alpha}2 \cos\frac{\alpha}2\right) = a^2\left(2\sin \frac{\alpha}2 + \frac12 \sin \alpha\right)$$

$$\frac{dA}{d\alpha} = a^2 \left(\cos \frac{\alpha}2 + \frac12 \cos \alpha \right)$$

Now set $\frac{dA}{d\alpha} = 0$ and solve for $\alpha$. Then check that the sign of the second derivative at $\alpha$ is negative.

Note that you don't need to assume $a=1$ anywhere. The $a$ disappears when taking $\frac{dA}{d\alpha} = 0$.

Hints on solving $\frac{dA}{d\alpha} = 0$:

$$a^2 \left(\cos \frac{\alpha}2 + \frac12 \cos \alpha \right) = 0 \implies \\ \cos \frac{\alpha}2 + \frac12 \left(2 \cos^2 \frac{\alpha}2 - 1\right) = 0 \implies \\ \cos \frac{\alpha}2 + \cos^2 \frac{\alpha}2 - \frac12 = 0$$

Now, set $x = \cos \frac{\alpha}2$ and solve $x + x^2 - \frac12 = 0$.

edited Oct 21 '13 at 12:12

answered Oct 21 '13 at 11:25

Daniel R

3,199

Thanks for nice nice picture.
so when I solve for $\alpha$ I got $\alpha = -2\arccos\frac{\alpha}{2}$. However my sources (facit) tells me that it should be $2\arccos\left(\frac{\sqrt{3}-1}{2}\right)$ So it's still wrong =(
– iveqy Oct 21 '13 at 11:50
@iveqy You're welcome! Check my edit, can you go from there? – Daniel R Oct 21 '13 at 12:12
Did you find the solution, @iveqy? – Daniel R Oct 22 '13 at 09:03
1

Yes finally, thank you! – iveqy Oct 22 '13 at 19:54

Maximize area depending on angle

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