1. We need to maximize $p+q$ subject to the condition $p^2+q^2=1,$ where $p,q>0$.

2. Use the identity $(p+q)^2=p^2+q^2+2pq.$ Since $p^2+q^2=1$, we get $(p+q)^2=1+2pq.$

3. Now, for fixed $p^2+q^2=1$, the product $pq$ is maximized when $p=q.$ This can also be seen from $(p-q)^2\ge 0 \implies p^2+q^2\ge 2pq \implies 1\ge 2pq \implies pq\le \frac12.$

4. Therefore, $(p+q)^2=1+2pq \le 1+2\cdot \frac12=2.$ So, $p+q\le \sqrt{2}.$

5. Equality occurs when $p=q.$ Using $p^2+q^2=1$, $2p^2=1 \implies p=q=\frac{1}{\sqrt{2}}.$ Hence the maximum value is $p+q=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}.$

6. Checking options:

• A: $\frac12$ — incorrect
• B: $\frac{1}{\sqrt2}$ — incorrect
• C: $\sqrt2$ — correct
• D: $2$ — incorrect

Therefore, the correct answer is Option C.