Key Concepts and Formulas

• Trigonometric Substitution: For equations of the form $x^2 + y^2 = r^2$, we can substitute $x = r\cos\theta$ and $y = r\sin\theta$.
• $R\sin(\theta + \alpha)$ Form: $a\cos\theta + b\sin\theta = R\sin(\theta + \alpha)$, where $R = \sqrt{a^2 + b^2}$ and $\tan\alpha = \frac{a}{b}$.
• Maximum value of $\sin(\theta)$: The maximum value of $\sin(\theta)$ is 1.

Step-by-Step Solution

Step 1: Understand the problem and apply trigonometric substitution

We are given $p^2 + q^2 = 1$ and $p, q > 0$. We want to find the maximum value of $p+q$. Since $p^2 + q^2 = 1$, we can use the trigonometric substitution $p = \cos\theta$ and $q = \sin\theta$.

Why? This substitution automatically satisfies the given equation $p^2 + q^2 = 1$. It also expresses $p$ and $q$ in terms of a single variable $\theta$, simplifying the maximization problem.

Since $p > 0$ and $q > 0$, we have $\cos\theta > 0$ and $\sin\theta > 0$. This implies that $0 < \theta < \frac{\pi}{2}$.

Step 2: Express $p+q$ in terms of $\theta$

We want to maximize $p+q$. Substituting $p = \cos\theta$ and $q = \sin\theta$, we get:

$p+q = \cos\theta + \sin\theta$

Why? This step converts the expression to be maximized into a function of a single variable $\theta$.

Step 3: Find the maximum value of $\cos\theta + \sin\theta$

We can rewrite $\cos\theta + \sin\theta$ in the form $R\sin(\theta + \alpha)$. Here, $a = 1$ and $b = 1$. Therefore, $R = \sqrt{1^2 + 1^2} = \sqrt{2}$. So, $\cos\theta + \sin\theta = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta\right) = \sqrt{2}\left(\sin\frac{\pi}{4}\cos\theta + \cos\frac{\pi}{4}\sin\theta\right) = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)$

Why? This transformation allows us to use the fact that the maximum value of $\sin x$ is 1.

Since $0 < \theta < \frac{\pi}{2}$, we have $\frac{\pi}{4} < \theta + \frac{\pi}{4} < \frac{3\pi}{4}$. The maximum value of $\sin\left(\theta + \frac{\pi}{4}\right)$ in this interval is 1, which occurs when $\theta + \frac{\pi}{4} = \frac{\pi}{2}$, or $\theta = \frac{\pi}{4}$.

Therefore, the maximum value of $\cos\theta + \sin\theta$ is $\sqrt{2} \cdot 1 = \sqrt{2}$.

Why? We found the maximum value of the trigonometric expression, which corresponds to the maximum value of $p+q$.

Step 4: State the maximum value of $p+q$

The maximum value of $p+q$ is $\sqrt{2}$.

Common Mistakes & Tips

• Remember to consider the constraint on $\theta$ imposed by the positivity of $p$ and $q$.
• Be careful when transforming trigonometric expressions; ensure you are using correct identities.
• The $R\sin(\theta+\alpha)$ method is generally quicker than using calculus to find the maximum/minimum.

Summary

We used trigonometric substitution to transform the problem of maximizing $p+q$ subject to $p^2 + q^2 = 1$ and $p, q > 0$ into maximizing $\cos\theta + \sin\theta$ for $0 < \theta < \frac{\pi}{2}$. We then rewrote $\cos\theta + \sin\theta$ as $\sqrt{2}\sin(\theta + \frac{\pi}{4})$ and found that its maximum value is $\sqrt{2}$.

The final answer is \boxed{\sqrt{2}}, which corresponds to option (C).