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A rectangle with sides on the positive $x$-axis and positive $y$-axis is inscribed in the circle of radius $1$ with center at the origin. If $A(x)$ is the area of this rectangle for $\frac 35 \le x \le \frac 45$, so $A\colon \big[\frac 35, \frac 45\big] \to \mathbb R$, find those $x \in \big[\frac 35, \frac 45\big]$ where $A$ attains a maximum value and where $A$ attains a minimum value, or say no such values exist.

A graph of the problem looks like a circle with radius one centered on the origin, with a rectangle inside of it. The top-right corner of the rectangle stays in contact with the circle no matter what length is chosen, i.e. the length and width are related in that changing the length requires the width to be changed so that the top-right corner stays touching.

Here is what I have so far:

  • Since $A$ is base $\cdot$ height, and the problem is wanting to find the base ($x$) that optimizes, the height needs to be expressed in terms of the base. Since the rectangle is in a circle, the height is going to be radius $-$ width, radius is 1, so that is $1 - $ width.
  • So, $A = w(1 - w)$, or $A = w - w^2$, and $A'(x) = 1 - 2w$
  • The root of the derivative $A'(x) = 0$ is, by my calculation, $\frac 12$
  • $\frac 12$ is outside the interval

Am I correct in concluding that the maximum and minimum area for any $w$ on the interval are $A({\scriptstyle\frac 35})$ and $A({\scriptstyle\frac 45})$, respectively?


I apologize, I'm having trouble with MathJaX here.

Corrected, $h = \sqrt{1 - w^2}$. So $A(b) = b \cdot \sqrt{1 - b^2}$. $A'(b) = \frac{1 - 2b^2}{\sqrt{1 - b^2}}$.

I still get the roots as $-\sqrt{\frac 12}$ and $+\sqrt{\frac12}$, can anyone see what I may be doing wrong?

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Your formula for the height is not correct. If $a, b$ are the sides of the rectangle then $a^2+b^2 = 1$, by Pythagoras, since the radius is the diagonal of the rectangle. – user20266 Mar 28 '12 at 8:49
@Thomas: The diagonal of the rectangle is the diameter, so $a^2 + b^2 = 2$. – martini Mar 28 '12 at 9:24
$A(3/5)$ and $A(4/5)$ are equal because of symmetry: $(3/5)^2+(4/5)^2=1$ – Henry Mar 28 '12 at 9:26
@martini: I think the diagonal of the rectangle is the radius – Henry Mar 28 '12 at 9:27
@Henry Thx, you and Thomas are right. I didn't read carefully. – martini Mar 28 '12 at 9:31

Assume you're given a rectangle $\mathcal{R}$ and a couple of vertices, say $P_1=(x_1,y_1),\ P_2=(x_2,y_2)$, which are opposite. Then the area of $\mathcal{R}$ is given by $A=(x_2-x_1)(y_2-y_1)$.

In your problem, your rectangle has a vertex in $O=(0,0)$ and the vertex opposite to $O$ in a variable point, say $P=(x,y)$, hence the area of your rectangle is $A=xy$.

Now there are two possible ways to solve your problem:

  1. (Cartesian coordinates) The variable point $P$ lies on an arc of the unit circumference, precisely on an arc placed into the first quardant (i.e. in the set ${(x,y)\in \mathbb{R}^2:\ x>0,y>0}$); since the unit circumference has equation $x^2 + y^2 = 1$, it is clear that the $y$-coordinate of the variable point $P$ can be written as $\sqrt{1-x^2}$. Therefore, you get the area function: $$A(x)=x\sqrt{1-x^2}\; ,$$ which you want to optimize in $[3/5, 4/5]$. Observe that $A$ is a continuous function of $x$ and that $x$ lies into a closed and bounded interval: thus Weierstrass theorem applies and it yields that $A$ attains a maximum and a minimum value in $[3/5, 4/5]$. In order to find such values, since $A$ has continuous derivative, you compute: $$A^\prime (x) = \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}} = \frac{1-2x^2}{\sqrt{1-x^2}}$$ and see that $A^\prime (x)\geq 0$ iff $3/5\leq x\leq 1/\sqrt{2}$ and $A^\prime (x)\leq 0$ otherwise; therefore $A$ is strictly increasing in $[3/5 , 1/\sqrt{2}]$ and strictly decreasing in $[1/\sqrt{2}, 4/5]$. As a consequence, you obtain: $$A_\max = A(1/\sqrt{2}) = 1/2\; ,$$ and: $$A_\min = \min \{ A(3/5) ,A(4/5)\} = 12/25\; .$$

  2. (Polar coordinates) By the very definition of sine and cosine, you obtain that the variable point $P$ has coordinate $(\cos \theta ,\sin \theta)$, where $\theta \in ]0,\pi/2[$ is the angle between the positive $x$-axis and the radius $OP$. Hence the area function to be optimized rewrites: $$A(\theta) = \cos \theta\ \sin \theta = \frac{1}{2} \sin 2\theta\; .$$ Moreover, observe that $P$ lies on the arc of unit circumference having extrema $(3/5,4/5)$, $(4/5, 3/5)$ iff $\theta \in [\arcsin (3/5), \arcsin (4/5)]$, thus you have to optimize $A(\theta)$ in the latter interval, which is possible because of the Weierstrass theorem. As before, you compute: $$A^\prime (\theta) = \cos 2\theta\; ,$$ hence $A^\prime (\theta)\geq 0$ iff $\theta \in [\arcsin (3/5), \pi/4]$ and $A^\prime (\theta)\leq 0$ otherwise; hence $A$ strictly increases in $[\arcsin (3/5),\pi/4]$ and strictly decreases in $[\pi/4, \arcsin (4/5)]$. Therefore: $$A_\max = A(\pi/4) = 1/2$$ and: $$A_\min = \min \{ A(\arcsin (3/5)), A(\arcsin (4/5))\}\; ;$$ but: $$A(\arcsin (3/5)) = \cos \arcsin (3/5)\ \sin \arcsin (3/5) = 4/5\ 3/5=12/25$$ (because $\cos \arcsin x = \sqrt{1-x^2}$ and $\sin \arcsin x = x$ when $0\leq x\leq 1$) and similarly: $$A(\arcsin (4/5)) = 12/25\; ,$$ hence: $$A_\min = 12/25\; .$$

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