Mathematical Enigma - XXI

Method 1: For a semicircle of radius $r$ , draw a radial line from its center, (situated at the origin), to a point $A$ on the circle in the first quadrant such that the angle between the $x$ -axis and the radial line is $\theta \gt 0$ . The coordinates of $A$ are then $(r\cos(\theta), r\sin(\theta))$ . Going clockwise, the rectangle in question will then have coordinates $B(r\cos(\theta),0), C(-r\cos(\theta),0)$ and $D(-r\cos(\theta),r\sin(\theta)).$

The perimeter $P$ of rectangle $ABCD$ will then be

$P = 4r\cos(\theta) + 2r\sin(\theta) = 2\sqrt{5}r\left(\dfrac{2}{\sqrt{5}}\cos(\theta) + \dfrac{1}{\sqrt{5}}\sin(\theta)\right) = 2\sqrt{5}r\sin(\theta + \alpha)$ ,

where $\alpha = \sin^{-1}(\frac{2}{\sqrt{5}})$ . Choosing $\theta = \frac{\pi}{2} - \alpha$ allows us to maximize $P$ as $2\sqrt{5}r$ , which for $r = \sqrt{5}$ is $\boxed{10}$ .

Method 2: Let the coordinates of $A$ this time be $(x,y) = (x,\sqrt{r^{2} - x^{2}})$ , since the equation of the semicircle is $x^{2} + y^{2} = r^{2}$ . Then the perimeter $P(x)$ will be

$P(x) = 4x + 2y = 4x + 2\sqrt{r^{2} - x^{2}}$ .

Any critical points will occur when $\dfrac{dP}{dx} = 4 - \dfrac{2x}{\sqrt{r^{2} - x^{2}}} = 0$

$\Longrightarrow 4\sqrt{r^{2} - x^{2}} = 2x \Longrightarrow 4(r^{2} - x^{2}) = x^{2} \Longrightarrow 4r^{2} = 5x^{2} \Longrightarrow x = \dfrac{2r}{\sqrt{5}}$ ,

for which $P = \dfrac{8r}{\sqrt{5}} + 2\sqrt{r^{2} - \dfrac{4r^{2}}{5}} = \dfrac{8r}{\sqrt{5}} + \dfrac{2r}{\sqrt{5}} = 2\sqrt{5}r$ .

As $P(0) = 2r \lt 2\sqrt{5}r$ we can rest assured that $P$ will achieve a maximum at the critical point $x = 2\sqrt{5}r$ , which again for $r = \sqrt{5}$ comes out to $\boxed{10}$ .

Calc solution which is dumb b/c this is a geo problem.

Let $x$ =length of rectangle and width is $√(5-x^2)$ Perimeter= $2*x*√(5-x^2)$ Taking derivative and setting to 0 $4-\frac{2 x}{\sqrt(5-x^2)}$ , you get $x=2$ Thus plugging $x=2$ you arrive at perimeter= $10$

Let the origin be the center of the semicircle. Let (x,y) be the top right corner of the rectangle.
Then perimeter P=4x+2y. For extrema $\dfrac{dP}{dx}=4+2*\dfrac{dy}{dx}=0,~\\ \implies ~\color{#3D99F6}{\dfrac{dy}{dx}=-2.} \\ x^2+y^2=5, ~~\therefore ~x+y\dfrac{dy}{dx}=0, \implies ~x=2y.\\ \therefore ~ ~ the ~ semicircle~ equation, ~\{2y\}^2+y^2=5,~ implies ~y=\pm 1, ~substituting,~ x =\pm2.$

Since we can not have y= - 1, and the minimum perimeter is 0, the rectangle with vertices (-2,1),(-2,0),(2,0) and (2,1) gives maximum perimeter which is 1+4+1+4=10.

Let $a$ be the length of the side of rectangle $ABCD$ which lies on the base of the semicircle, $b$ be the length of the side of the rectangle $ABCD$ adjacent to the previously mentioned side and $p$ be the perimeter of rectangle $ABCD$ .

By Cauchy's Inequality:

$(\frac{a^2}{4}+b^2)(16+4) \geq (2a+2b)^2$

Note that by Pythagoras' Theorem, we have:

$\frac{a^2}{4}+b^2=5$

Also,

$p=2*(a+b)$

So,

$5*20\geq p^2$

$10\geq p$

So the maximum of the perimeter of rectangle $ABCD$ is $10$ .

Equality when $\frac{\frac{a^2}{4}}{16}=\frac{b^2}{4}\rightarrow a=4, b=1$ .

Taking the centre of semicircle as origin we write the polar coordinates same as that of Brian Charlesworth. Since now P =(2×rcosx + rsinx)×2 we have the max of acosy + bsiny as +sqrt(a^2+b^2) Pmax = sqrt(5)×r×2=5×2=10 HENCE PROVED