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A decorative garden is to have the shape of a circular sector of radius $r$ and central angle $\theta$. If the perimeter is fxed in advance, what value of $\theta$ will maximize the area of the garden?

I've given it a go, but I don't think I really understand the problem. Here's my try: the area of a sector is $A=\frac{1}{2}r^2 θ$. The perimeter being fixed means that we can treat $r$ as a constant. We need to find the value of $\theta\in[0,2\pi]$ that maximizes $A$, but since $A$ is really just a line with a positive slope, that value is $2\pi$. Needless to say that isn't the right answer according to the solution manual. So, what am I doing wrong? Thanks.

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  • $\begingroup$ the area is $r^2\theta$ $\endgroup$ – Martín Vacas Vignolo Jan 2 '18 at 19:31
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    $\begingroup$ The radius is not fixed; $P = 2r + r\theta$ is. $\endgroup$ – N. F. Taussig Jan 2 '18 at 19:32
  • $\begingroup$ @N.F.Taussig I see. So, I know that the perimeter of the circle is $2 \pi r$ and that the lenght of the subtended arc to $\theta$ is $r\theta$. I'm a bit lost now. Can you give me any hint on how to proceed from here? Thanks. $\endgroup$ – Sasaki Jan 2 '18 at 19:45
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    $\begingroup$ $2r + r \theta = P$ is a constant. You want to maximize $A = \frac{1}{2}r^2 \theta$. Can you express $A$ as a function of just $r$ but not $\theta$ (perhaps using the unknown constant $P$)? If so, then you just need to maximize a function of one variable. $\endgroup$ – John Hughes Jan 2 '18 at 19:47
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Let $P$ be the perimeter and $S$ be an area of the garden.

Thus, $2r+\theta r=P$, which gives $r=\frac{P}{2+\theta}$ and by AM-GM we obtain: $$S=\frac{1}{2}r^2\theta=\frac{P^2}{2}\cdot\frac{\theta}{4+4\theta+\theta^2}=\frac{P^2}{2}\cdot\frac{1}{\frac{4}{\theta}+\theta+4}\leq$$ $$\leq\frac{P^2}{2}\cdot\frac{1}{2\sqrt{\frac{4}{\theta}\cdot\theta}+4}=\frac{P^2}{16}.$$ The equality occurs for $\theta=2$.

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  • Compute the perimeter $P$: $P=2r+\theta r =r(2+ \theta)$
  • Solve for r and replace r in the formula for the area: $A=\frac{1}{2} \theta (\frac{P}{2+\theta})^2$
  • Find the $\theta$ which maximize the area: $\frac{d A}{d \theta}=0$. $\theta_{max}=2$.

If you are working with angles measured in degrees, instead of in radians, then you'll need to include a conversion factor.

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For sector shaped area $A$ radius $r$

Lagrangian $A- \lambda p $

$$ A= \frac12 r^2 \theta,\quad p = 2r + r \theta $$

$$ \dfrac{A_r}{A_\theta}=\dfrac{p_r}{p_\theta}$$

$$ \dfrac{r \theta}{r^2/2}=\dfrac{2+\theta}{r}$$

$$ \rightarrow \theta =2, A=r^2, p=4r$$

and $$ p^2 = 16 A $$

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