Question-49
The sum of two positive numbers is \(\displaystyle 16\). What is the smallest positive value of the sum of their squares?
\(128\)
Let the two numbers be \(\displaystyle x,16-x\). We need to solve:
\[ \begin{equation*} \underset{x}{\max} \ x^{2} +( 16-x)^{2} \end{equation*} \]
Let \(\displaystyle f( x) =x^{2} +( 16-x)^{2}\). Then:
\[ \begin{equation*} f^{\prime }( x) =2x-2( 16-x) =4x-32 \end{equation*} \]
We have a critical point at \(\displaystyle x=8\). \(\displaystyle f^{\prime \prime }( 8) =4 >0\), which suggests that \(\displaystyle x=8\) is a local minimum for \(\displaystyle f\). Since \(\displaystyle f\) is bounded in the interval \(\displaystyle [ 0,16]\) and is continuous, we also check for the boundaries. \(\displaystyle f( 0) =f( 16) =16^{2}\) which is clearly greater than \(\displaystyle f( 8) =2\times 8^{2}\). Hence, \(\displaystyle x=8\) corresponds to a global minimum of \(\displaystyle f\) in the interval \(\displaystyle [ 0,16]\). The required value is \(\displaystyle \boxed{128}\).