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

Question

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!

Answer 1

You have dropped an $x$ in calculating your derivative. By applying the product rule: $$\begin{align}A'(x) &= 2x\left(\frac{1}{2}(r^2-x^2)^{-1/2}(-2\color{red}{x})\right) + 2\sqrt{r^2-x^2}\\ &= \frac{-2x^{\color{red}{2}}}{\sqrt{r^2-x^2}} + 2\sqrt{r^2-x^2}\end{align}$$

Answer 2

hint :$x\sqrt{r^2-x^2}=\sqrt{x^2(r^2-x^2)}\le \dfrac{x^2+(r^2-x^2)}{2}=\dfrac{r^2}{2}$

Answer 3

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 $\theta=\frac{\pi}{4}$.

Answer 4

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

Answer 5

Eqn of circle: x^2 + y^2 = r^2

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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)

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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)

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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)

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finding y:

==>y^2 = r^2 - x^2

==>y^2 = r^2 - (r^2)/2

==>y=r/sqrt(2)