Question

The two positive numbers, whose sum is $15$ and the sum of whose squares is minimum are

• A. $\frac{15}{2}$, $\frac{13}{2}$
• B. $\frac{13}{2}$, $\frac{11}{2}$
• C. $\frac{15}{2}$, $\frac{15}{2}$
• D. $\frac{17}{2}$, $\frac{13}{2}$

Answer 1

Answer: (C)

Let one of the numbers be $x$. Then, the other number is $(15-x)$. Let $S(x)$ denote the sum of the squares of these numbers. Then,
$S\left(x\right)=x^2+{\left(15-x\right)}^2$
$=2x^2-30x+225$
$\Rightarrow S^{\prime }\left(x\right)=4x-30$
$\Rightarrow S^{\prime \prime }\left(x\right)=4$
Now, $S^{\prime }\left(x\right)=0$ gives $x=\frac{15}{2}$.
Also, $S^{\prime \prime }\left(\frac{15}{2}\right)=4>0$.
Therefore, by second derivative test, $x=\frac{15}{2}$ is the point of local minima of $S$. Hence, the sum of squares of numbers is minimum when the numbers are $\frac{15}{2}$ and $15-\frac{15}{2}=\frac{15}{2}$.