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Consider this image: enter image description here

A rectangle is inscribed in a semicircle and the radius is 1. The bas of the rectangle is x. Write an expression for the rectangle perimeter and determine the value of x that gives the highest possible perimeter. Also, what is the highest perimeter? Well... I fooled around with the unit circle, but still no progress. I've found several instructions on how to find the area, but no luck with perimeter.

I've been trying to solve this one for days now, I have completely giving upp solving it by my self. If someone could provide me with both answers & solutions, I would be happier than a child in a candy store :)

Could any kind soul help me out? :)

EDIT:

Please read new update:

http://imgur.com/a/adATx

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    $\begingroup$ Have you tried involving trig functions? It makes the problem much easier. $\endgroup$ – David Quinn Jan 2 '17 at 22:35
  • $\begingroup$ Try derivatives $\endgroup$ – user398748 Jan 2 '17 at 22:46
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    $\begingroup$ Next time you ask a question, rather than linking to an image of your calculations, why not put the actual words and formulas (not pictures of them) in the question? Use MathJax to format the formulas; see meta.math.stackexchange.com/questions/5020/… $\endgroup$ – David K Jan 3 '17 at 19:27
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I think you have come till this equation: $$P=2x+\sqrt{4-x^2}$$

Further, $$\frac{dP}{dx}= 2-\frac{x}{\sqrt{4-x^2}}$$ Set $\frac{dP}{dx}=0$ for finding the maximum of P.

So, $$2=\frac{x}{\sqrt{4-x^2}}$$ $x=\sqrt{\frac{16}{5}}$ is the value of x for maximum value of $P$ (which gives the maximum perimeter).

Therefore, the perimeter of the rectangle when $x=\sqrt{\frac{16}{5}}$ is $P=2\sqrt{5}$

(by substituting the value of x in the first equation).

Finally, $$x=\frac{4}{\sqrt{5}}$$

and

$perimeter(max)=2\sqrt{5}$

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Hint: calculate the height with pitagoras (hyp=1)

Edit: the height is $h=\sqrt{1-x^2/4}$. Now, the perimeter is $p(x)=2x+2 \sqrt{1-x^2/4}$.

$p'(x)=2+(1-x^2/4)^{-1/2}(-x/2)=0$ iff $1-x^2/4=x^2/16$ iff $x^2=16/5$

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  • $\begingroup$ thanks, but done that already. $\endgroup$ – T.Bill Jan 2 '17 at 22:33
  • $\begingroup$ If you have already found $h$ in terms of $x$ using the Pythagorean theorem, then you should be able to express the perimeter $P$ in terms of $x$ and $h$, then substitute your value for $h$ in terms of $x$. Then you will have $P$ as a function of $x$. Then you will have to set the derivative equal to $0$ to find the optimum value of $x$. $\endgroup$ – John Wayland Bales Jan 2 '17 at 22:47
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$P=2x+2h$

By definition $x=r\cos(\theta) \implies x=\cos(\theta)$

$P=2\cos(\theta)+2h$

By pythagoras $(\frac{x}{2})^2+h^2=1$

$h^2=1-\frac{x^2}{4}$

$h^2=\frac{1}{4}(4-x^2)$

$h^2=\frac{1}{4}(4-\cos^2(\theta)$)

$h=\frac{1}{2}\sqrt{4-\cos^2(\theta)}$

$P=2x+2h$

$P=2\cos(\theta)+2(\frac{1}{2}\sqrt{4-\cos^2(\theta)})$

$P=2\cos(\theta)+\sqrt{4-\cos^2(\theta)}$

Notice that the $-1\leq\cos(\theta) \leq1$ and $0 \leq \cos^2(\theta) \leq 1$

So max value is $P(\theta)_{max}=2+\sqrt{4-1}=2+\sqrt{3}$

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The perimeter is given by,

$$2h+2x=2(h+x)$$

We need to maximize $2(h+x)=2\sqrt{1-x^2/4}+2x=\sqrt{4-x^2}+2x$ by pythag. Thm. Obviously $x \in [0,2]$.

Let $x=2 \sin (\theta)$, restrict $\theta \in [0, \frac{\pi}{2}]$.

We have,

$$\sqrt{4-x^2}+2x=2\cos(\theta)+4\sin (\theta) = \sqrt{2^2+4^2} \cos (\theta-\omega)$$

Where $\omega=\arctan (\frac{4}{2})$ and $\omega \in [0,\frac{\pi}{2}]$

We can get $\theta-\omega=0$, so the maximum is $\sqrt{2^2+4^2}$

Identities used include $\sin^2 (\theta)+ \cos^2 (\theta)=1$ and:

Given, $a,b>0$: $a\cos (\theta)+b \sin (\theta)=\sqrt{a^2+b^2}\cos( \theta-\omega)$

Where $\omega$ is the angle in the right triangle with legs of length $a$ and $b$ which is opposite to $b$.

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Call the angle between the bottom of the semicircle and the radius to the vertex of the rectangle $\theta$. Then the height of the rectangle is $h = \sin \theta$ and the length is $x = 2\cos\theta$, making the perimeter $P = 2(x+h) = 4\cos\theta + 2 \sin\theta$. Using the sum identity, we can write this as

$$ P = 4\cos\theta + 2 \sin\theta = 2\sqrt{5}\left(\frac{2}{\sqrt{5}}\cos\theta + \frac{1}{\sqrt{5}}\sin\theta\right) = 2\sqrt{5}\cos\left[\theta - \cos^{-1}\left(\frac{2}{\sqrt{5}}\right)\right] $$

From this you can easily find the maximum value of $P$ and the corresponding value of $x = 2\cos\theta$.

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