The number 10 is divided into two parts such | vedclass

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(A) Let the two parts of $10$ be $x$ and $(10-x)$.Define the function $f(x) = 2x + (10-x)^2$.Expanding the function: $f(x) = 2x + 100 - 20x + x^2 = x^

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$9, 1$

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(A) Let the two parts of $10$ be $x$ and $(10-x)$.Define the function $f(x) = 2x + (10-x)^2$.Expanding the function: $f(x) = 2x + 100 - 20x + x^2 = x^2 - 18x + 100$.To find the minimum,find the first derivative: $f'(x) = 2x - 18$.Set $f'(x) = 0$ to find the critical point: $2x - 18 = 0 \implies x = 9$.Find the second derivative: $f''(x) = 2$.Since $f''(9) = 2 > 0$,the function has a minimum at $x = 9$.The first part is $x = 9$ and the second part is $10 - 9 = 1$.Therefore,the numbers are $9$ and $1$.

The number $10$ is divided into two parts such that the sum of double of the first part and the square of the second part is minimum. The two parts are respectively:

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