AIEEE 2007 (Offline)Question 1
If $p$ and $q$ are positive real numbers such that $p^{2}+q^{2}=1$, then the maximum value of $(p+q)$ is
Differentiation and its Application
Solution
Concept. Compare $(p+q)^2$ with $2(p^2+q^2)$. Formulas used. (1) $(p+q)^2\le2(p^2+q^2)$, since $2(p^2+q^2)-(p+q)^2=(p-q)^2\ge0$Step 1 - Bound. By (1), $(p+q)^2\le2$, so $p+q\le\sqrt2$. Step 2 - Equality. At $p=q=\frac{1}{\sqrt2}$, $p+q=\sqrt2$. Maximum $\sqrt2$. Answer: (C)