Review question

# As $\theta$ varies, what's the largest this ratio of areas can be? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9770

## Solution

1. Let $a > 0$. Sketch the graph of

$y=\dfrac{a+x}{a-x} \quad \text{for} \quad -a < x < a.$

Now $y=\dfrac{a+x}{a-x} = \dfrac{2a-(a-x)}{a-x} = \dfrac{2a}{a-x}-1$, so as $x$ increases from $-a$ towards $a, y$ increases from $0$ towards $\infty$. Also $(0,1)$ is on the curve.

If we use calculus, $\dfrac{dy}{dx} = \dfrac{2a}{(a-x)^2} > 0$, and so this function is always increasing (with a discontinuity at $x = a$).

1. Let $0 < \theta < \dfrac{\pi}{2}$. In the diagram below is the half-disc given by $x^2 + y^2 \leq 1$ and $y \geq 0$.

The shaded region $A$ consists of those points with $-\cos \theta \leq x \leq \sin \theta$. The region $B$ is the remainder of the half-disc.

Find the area of $A$.

We can see from the diagram that A consists of a quarter of the circle with two additional congruent triangles.

Thus the area $A = \dfrac{\pi}{4}+2\left[\dfrac{1}{2}\sin\theta\cos\theta\right] = \dfrac{\pi}{4}+\sin\theta\cos\theta$.

1. Assuming only that $\sin^2\theta + \cos^2\theta = 1$, show that $\sin\theta\cos\theta \leq \dfrac{1}{2}$.

We know that $0\leq (\sin\theta-\cos\theta)^2 = \sin^2\theta + \cos^2\theta - 2\sin\theta\cos\theta.$

Thus $2\sin\theta\cos\theta \leq \sin^2\theta + \cos^2\theta = 1 \implies \sin\theta\cos\theta \leq \dfrac{1}{2}.$

1. What is the largest $\dfrac{\text{area of } A}{\text{area of } B}$ can be, as $\theta$ varies?

We know $\dfrac{\text{area of } A}{\text{area of } B} = \dfrac{\dfrac{\pi}{4}+\sin\theta\cos\theta}{\dfrac{\pi}{2}-\left(\dfrac{\pi}{4}+\sin\theta\cos\theta\right)}= \dfrac{\dfrac{\pi}{4}+\sin\theta\cos\theta}{\dfrac{\pi}{4}-\sin\theta\cos\theta}$.

Now from part (i), we know that $\dfrac{\dfrac{\pi}{4}+\sin\theta\cos\theta}{\dfrac{\pi}{4}-\sin\theta\cos\theta}$ increases as $\sin\theta\cos\theta$ increases.

But the maximum value $\sin\theta\cos\theta$ can take is $\dfrac{1}{2}$, and so the maximum value of $\dfrac{\dfrac{\pi}{4}+\sin\theta\cos\theta}{\dfrac{\pi}{4}-\sin\theta\cos\theta}$ is $\dfrac{\dfrac{\pi}{4}+\dfrac{1}{2}}{\dfrac{\pi}{4}-\dfrac{1}{2}} = \dfrac{\pi+2}{\pi-2}$.