10 is divided into two parts such that the sum of double of the first and square of the other is minimum,…
10 is divided into two parts such that the sum of double of the first and square of the other is minimum, then the numbers are respectively
9,1
8,2
6,4
7,3
Solution
Let the two parts of 10 be $x$ and $(10-x)$.
$f(x)=2(10-x)+x^2=x^2-2 x+20$
$f^{\prime}(x)=2 x-2$ and when $f^{\prime}(x)=0$, we get $x=1$
$\mathrm{f}^{\prime \prime}(\mathrm{x})=2>0$
$\therefore \mathrm{f}(\mathrm{x})$ is minimum at $\mathrm{x}=1$.
Thus the parts are 1,9 .