The two parts of 100 for which the sum of dou | vedclass

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(B) Let the two parts be $x$ and $(100 - x)$.Let the first part be $(100 - x)$ and the second part be $x$.The function to minimize is $f(x) = 2(100 -

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$99, 1$

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(B) Let the two parts be $x$ and $(100 - x)$.Let the first part be $(100 - x)$ and the second part be $x$.The function to minimize is $f(x) = 2(100 - x) + x^2$.$f(x) = x^2 - 2x + 200$.To find the critical points,we find the derivative $f'(x) = 2x - 2$.Setting $f'(x) = 0$,we get $2x - 2 = 0$,which implies $x = 1$.Now,we check the second derivative: $f''(x) = 2$.Since $f''(x) = 2 > 0$,the function has a local minimum at $x = 1$.Thus,the second part is $1$ and the first part is $100 - 1 = 99$.The two parts are $99$ and $1$.

The two parts of $100$ for which the sum of double of the first part and the square of the second part is minimum,are

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