A rectangle is inscribed in a circle sector. The top two corners of the rectangle lies on the radius of the circle sector and the bottom two corners lie on the arc of the circle sector.

enter image description here The radius is 2 and angle is $\frac{2\pi}{3}$.

Find the maximum area of the rectangle that has these properties and are located exactly like this using calculus-based optimization.

Attempted solution:

Let the shorter side of the rectangle be labelled x and the longer be labelled y. The distance from the center of the circle to the top of the rectangle is b.

Now, create a dashed-line triangle from the center of the circle to the bottom right of the rectangle. Call the top angle $\alpha$.

From this, we can get a value for $\frac{y}{2}$, since this is the side opposite the angle:

$$\sin \alpha = \frac{\frac{y}{2}}{2} \Rightarrow y = 4 \sin \alpha$$

Using the same triangle but with the adjacent side x + b:

$$\cos \alpha = \frac{x+b}{2} \Rightarrow x = 2 \cos \alpha - b$$

Then we use half of the top triangle above the rectangle to get a value for b in terms of y:

$$\tan \frac{2\pi}{6} = \frac{\frac{y}{2}}{b} \Rightarrow b = \frac{\frac{1}{2}y}{\tan \frac{\pi}{3}}$$

Combining these to get a value for x:

$$x = 2 \cos \alpha - b = 2 \cos \alpha - \frac{\frac{1}{2}y}{\tan \frac{\pi}{3}} = 2 \cos \alpha - \frac{2 \sin \alpha}{\tan \frac{\pi}{3}}$$

The area becomes:

$$A(\alpha) = yx = 4 \sin \alpha (2 \cos \alpha - \frac{2 \sin \alpha}{\tan \frac{\pi}{3}})$$

Taking the derivative:

$$A'(\alpha) = 8(-sin^2 \alpha + cos^2 \alpha - 2 \sin \alpha \cos \alpha)$$

Setting to zero:

$$A'(\alpha) = 0 \Rightarrow a = \frac{\pi}{2} - \frac{3\pi}{8}$$

Putting this into the formula for A:

$$A( \frac{\pi}{2} - \frac{3\pi}{8}) = 4(\sqrt{2} -1)$$

However, this does not match the expected answer of $\frac{4}{\sqrt{3}}$.

What went wrong? What are some productive ways to finish this off?

  • 1
    $\begingroup$ When you differentiated $A(\alpha)$, you dropped the $\tan \frac{\pi}{3} = \sqrt{3}$. $\endgroup$ – andy87 Jul 27 at 12:07
  • $\begingroup$ There is a nice solution without analysis. $\endgroup$ – Michael Rozenberg Jul 27 at 13:59
  • $\begingroup$ @MichaelRozenberg Would you be willing to give a hint for the nice solution without analysis? Thanks! $\endgroup$ – andy87 Jul 27 at 17:28
  • $\begingroup$ @andy87 If $CD$ is an altitude of $\Delta ABC$ with $\measuredangle ACB=90^{\circ}$, so $CD^2=AD\cdot DB.$ $\endgroup$ – Michael Rozenberg Jul 27 at 17:30

$A'(\alpha) = 8(-\sin^{2}{\alpha} + \cos^{2}{\alpha} - \frac{2}{\sqrt{3}} \sin{\alpha}\cos{\alpha})$. Setting this to zero gives the family of solutions $\alpha = \frac{1}{6}(3 \pi n + \pi)$, $n \in \mathbb{Z}$. Taking $n = 0$ gives $A(\pi/6) = \frac{4}{\sqrt{3}}$.


I got $$A(\alpha)=\frac{\sin(2\alpha)}{2}-\frac{\sin^2(\alpha)}{\tan(60^{\circ})}$$ so $$A'(\alpha)=8\left(\cos(2\alpha)-\frac{\sin(2\alpha)}{\tan(60^{\circ})}\right)$$ and $$A'(\alpha)=0$$ if $$\tan(2\alpha)=\tan(60^{\circ})$$


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