Exercise 6.3 · Q19: Of all rectangles inscribed in a circle, the square has the largest area

Basics you need

Solution

1. Every inscribed rectangle has a diameter as its diagonal

   Let the circle have centre O and radius $r$ (fixed). Let ABCD be any rectangle inscribed in it. The angle ABC is a right angle and B lies on the circle, so the chord AC on which it stands is a diameter (converse of "the angle in a semicircle is a right angle"). Hence

   $$AC = 2r.$$
   Another way to see it: the diagonals of a rectangle are equal and bisect each other, so their midpoint is the same distance from all four corners. The only point equidistant from three points of a circle is its centre, so the diagonals meet at O and each is a diameter.
   Circle theorem: the angle in a semicircle is a right angle, and a right angle inscribed in a circle stands on a diameter.

2. The relation between the sides, and the allowed range

   Let the sides be $AB = x$ and $BC = y$⁠. Triangle ABC has a right angle at B, so by Pythagoras

   $$x^2 + y^2 = (2r)^2 = 4r^2.$$

   Hence $y = \sqrt{4r^2 - x^2}$⁠, and a genuine rectangle needs $x > 0$ and $y > 0$⁠:

   $$0 < x < 2r.$$
   Conversely, every $x$ in $(0, 2r)$ gives an inscribed rectangle (draw any diameter as the diagonal and a right angle on it). So maximising over $0 < x < 2r$ covers all inscribed rectangles, which is what "of all the rectangles" requires.
   $y > 0$ means $4r^2 - x^2 > 0$⁠, that is $x < 2r$⁠. Basics: interval notation

3. The area as a function of $x$

   $$A(x) = xy = x\sqrt{4r^2 - x^2},\qquad 0 < x < 2r.$$
   Area of a rectangle = length × breadth; substituting for $y$ leaves one variable. Here $r$ is a fixed constant, not a variable.
   Area of a rectangle. Basics: building the function

4. Differentiate

   Product rule, with the chain rule for the square root:

   $$\begin{aligned} A'(x) &= \sqrt{4r^2 - x^2} + x\cdot\frac{-x}{\sqrt{4r^2 - x^2}} \\ &= \frac{(4r^2 - x^2) - x^2}{\sqrt{4r^2 - x^2}} \\ &= \frac{2(2r^2 - x^2)}{\sqrt{4r^2 - x^2}}. \end{aligned}$$
   To combine the two terms, write the first one over the common denominator $\sqrt{4r^2 - x^2}$⁠. On $0 < x < 2r$ the denominator is positive, so $A$ is differentiable on the whole range and its only critical points are where the numerator is 0.
   $\frac{d}{dx}\sqrt{u} = \frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$ with $u = 4r^2 - x^2$⁠, $\frac{du}{dx} = -2x$⁠. Basics: the chain rule

5. Critical point and sign of $A'$

   $A'(x) = 0 \iff x^2 = 2r^2 \iff x = \sqrt2\,r$ (the negative root is outside the range; $\sqrt2\,r < 2r$⁠, so it is inside). Since the denominator is positive, $A'(x)$ has the sign of $2r^2 - x^2$⁠:

   • for $0 < x < \sqrt2\,r$⁠: $x^2 < 2r^2$⁠, so $A'(x) > 0$ and $A$ is increasing;
   • for $\sqrt2\,r < x < 2r$⁠: $x^2 > 2r^2$⁠, so $A'(x) < 0$ and $A$ is decreasing.

   By the first derivative test (Theorem 3) $x = \sqrt2\,r$ is a point of local maxima, and the sign pattern on the whole range shows $A(x) < A(\sqrt2\,r)$ for every other inscribed rectangle.

   The first derivative test fits best here: the sign of $A'$ is visible from the factored form, whereas $A''$ is messy. The same sign information gives the global conclusion.
   First Derivative Test, Theorem 3, Section 6.4; Theorem 1, Section 6.3. Lesson: derivative tests

6. The largest rectangle is a square

   At $x = \sqrt2\,r$⁠:

   $$y = \sqrt{4r^2 - 2r^2} = \sqrt{2r^2} = \sqrt2\,r = x.$$

   The two adjacent sides are equal, so the rectangle of maximum area is a square, of side $\sqrt2\,r$ and area $2r^2$⁠. This is what was to be shown.

   A rectangle with two equal adjacent sides is a square. Its area $x \cdot y = \sqrt2\,r \cdot \sqrt2\,r = 2r^2$⁠.
   $\sqrt{2r^2} = \sqrt2\,r$ because $r > 0$⁠. Basics: area and volume formulas

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