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Q.
The two positive numbers, whose sum is $15$ and the sum of whose squares is minimum are
Application of Derivatives
Solution:
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'\left(x\right) = 4x - 30 $
$\Rightarrow S''\left(x\right) = 4$
Now, $S'\left(x\right) = 0$ gives $x = \frac{15}{2}$.
Also, $S''\left(\frac{15}{2}\right) = 4 > 0$.
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 $15 - \frac{15}{2} = \frac{15}{2}$.