Question

Find 2 +ve nos. whose sum is 15 and the sum of whose squares is min.

Answer 1

Let first number be x

Given sum of two positive number is 15

$1^{st}no+2^{nd}no=15$

$\to x+2^{nd}no=15$

$\to 2^{nd}=15-x$

let $S\left(x\right)$ be sum of squares of numbers

$S\left(x\right)=(1stno{)}^2+(2^{nd}no{)}^2$

$S\left(x\right)=x^2+(15-x{)}^2$

mininize $S\left(x\right)$

$S\left(x\right)=x^2+(15-x{)}^2$

$S^{\prime }\left(x\right)=2x+2(15-x)(-1)=2x-30+2x$

$S^{\prime }\left(x\right)=4x-30$

Putting $S^{\prime }\left(x\right)=0$

$\to 4x-30=0$

$\to x=\left(\frac{15}{2}\right)=7.5$

$S"\left(x\right)=d(\frac{(4x-30)}{dx}=4$

putting $x=7.5$ in $S"\left(x\right)$

$S"\left(7.5\right)=4>0,i.ex=7.5$ is local minima

$\therefore 1^{st}number=\frac{15}{2},2^{nd}no=\frac{15}{2}$