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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
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Verified Answer
The correct answer is:
9,1
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 .
$$
f(x)=2(10-x)+x^2=x^2-2 x+20
$$
$f^{\prime}(x)=2 x-2$ 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 .
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