# maximum area of a rectangle inscribed in a semi - circle with radius r.

A rectangle is inscribed in a semi circle with radius $r$ with one of its sides at the diameter of the semi circle. Find the dimensions of the rectangle so that its area is a maximum.

My Try:

Let length of the side be $x$, Then the length of the other side is $2\sqrt{r^2 -x^2}$, as shown in the image.

Then the area function is

$$A(x) = 2x\sqrt{r^2-x^2}$$

\begin{align}A'(x) &= 2\sqrt{r^2-x^2}-\frac{4x}{\sqrt{r^2-x^2}}\\ &=\frac{2}{\sqrt{r^2-x^2}} (r^2 - 2x -x^2)\end{align}

setting $A'(x) = 0$,

$$\implies x^2 +2x -r^2 = 0$$

Solving, I obtained:

$$x = -1 \pm \sqrt{1+r^2}$$

That however is not the correct answer, I cannot see where I've gone wrong? Can someone point out any errors and guide me the correct direction. I have a feeling that I have erred in the differentiation.

Also how do I show that area obtained is a maximum, because the double derivative test here is long and tedious.

Thanks!

• FYI for the future: All that you had left to do on LaTeX formatting was to wrap your code in $ signs. Good question, btw--I'll look into it. – apnorton Jul 20 '14 at 2:56 •$A'(x) = 2\sqrt{r^2-x^2}-\frac{2x^2}{\sqrt{r^2-x^2}}$– chenbai Jul 20 '14 at 3:03 • – Martin Sleziak Sep 11 '18 at 9:09 ## 8 Answers You have dropped an$xin calculating your derivative. By applying the product rule: \begin{align}A'(x) &= 2x\left(\frac{1}{2}(r^2-x^2)^{-1/2}(-2\color{red}{x})\right) + 2\sqrt{r^2-x^2}\\ &= \frac{-2x^{\color{red}{2}}}{\sqrt{r^2-x^2}} + 2\sqrt{r^2-x^2}\end{align} • Ah... I see, that was silly, thanks – GenAsis Jul 20 '14 at 3:06 • @GeniusAsis No problem. :) – apnorton Jul 20 '14 at 3:06 Let\theta$be the angle that the slanted red (?) line on the right makes with the horizontal. Then the height of the rectangle is$r\sin\theta$and the base is$2r\cos\theta$, for an area of$r^2\sin\theta\cos\theta$. This is$\frac{r^2}{2}\sin 2\theta$. But$\sin 2\theta$has a maximum value of$1$, at$\theta=\frac{\pi}{4}$. hint :$x\sqrt{r^2-x^2}=\sqrt{x^2(r^2-x^2)}\le \dfrac{x^2+(r^2-x^2)}{2}=\dfrac{r^2}{2}$# Eqn of circle: x^2 + y^2 = r^2 | # area of rectangle = a = 2xy # ==> a^2 = 4 x^2 y^2 # ==> a^2 = 4 x^2 (r^2 - x^2) # ==> a^2 = 4 (x^2 r^2 - x^4) | # differentiating: # ==> 2a da/dt = 4 (2x r^2 - 4x^3) # since, da/dt=0 # ==> 2 x r^2 = 4 x^3 # ==> x = r/sqrt(2) | # second derivative test: # ==>4 (2 r^2 - 12 x^2) # ==>8 r^2 - 24 r^2 # ==>(-16) r^2 # ==> -ve # therefore, maximum area for x= r/sqrt(2) | # finding y: # ==>y^2 = r^2 - x^2 # ==>y^2 = r^2 - (r^2)/2 # ==>y=r/sqrt(2) For convenience, think that the circle has its center at (0,0). We then consider the upper semicircle of x^2+y^2=a^2. (1) The area of the inscribed rectangle would be A=2xy dA/dx=(2x)'y+2x(dy/dx) =2y+2x(dy/dx) diff (1) (d/dx)(x^2+y^2)=(d/dx)(a^2) <=> 2x+2y(dy/dx)=0 <=> dy/dx = -x/y Now, dA/dx = 2y+2x(-x/y) <=> dA/dx = 2(y^2-x^2)/y <=>(1) dA/dx = 2(a^2-2x^2)/y dA/dx = 0 <=> a^2-2x^2 = 0 <=> x = a/sqrt(2) By elementary means: The maximum of$2x\sqrt{r^2-x^2}$is also that of$x^2\sqrt{r^2-x^2}$, by squaring, or that of$t(1-t)$, by setting$r^2t=x^2$. The expression$t(1-t)$describes a parabola, with roots at$t=0$and$t=1$, and by symmetry its vertex must be at$t=\dfrac12$. Hence the requested area, $$2\frac r{\sqrt 2}\sqrt{r^2-\frac{r^2}2}=r^2.$$ Doubling the semi-circle to obtain a full circle, we now have a rectangle inscribed in a circle, with area equal to twice the area of your starting rectangle. Letting$x$be either angle formed by its diagonals, we need to maximize $$A(x)=2r \sin x\,.$$ This has a maximum when the diagonals are orthogonal, i.e. when the rectangle is a square, with side$\sqrt 2\, r$. Hence, the dimensions of the original rectangle are$\sqrt2\,r$and$r/\sqrt2\,\$.

Equation of circle: $$x^2+y^2=r^2$$

$$x = \pm (r^2-y^2)$$

Thus, length of rectangle is $$x-(-x)=2x$$ and height is $$y$$.

Area $$A = 2xy$$. Maximizing A is equivalent to maximizing $$A^2$$

$$A^2 = 4x^2y^2 = 4(r^2-y^2)y^2=4r^2y^2-4y^4$$

Let, $$f(z)= 4r^2z-4z^2$$

Then, $$f'(z) = 4r^2-8z$$

Equating, $$f'(z)=0$$ we get, $$\mathbf{z=\frac12r^2}$$

Therefore, max. $$A^2$$ or max. $$f(z) = 4r^2\times \frac12r^2-4\times(\frac12r^2)^2 = 2r^4-r^4=r^4$$

Hence, $$\mathbf{max. A = \sqrt{r^4} = r^2}$$

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