Divide 10 into two parts such that the sum of | vedclass

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(D) Let $x$ and $y$ be the two parts of the number $10$. $	herefore x + y = 10 \implies y = 10 - x$ ... $(i)$ Let $A$ be the sum of double of the fir

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$(9, 1)$

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(D) Let $x$ and $y$ be the two parts of the number $10$. $	herefore x + y = 10 \implies y = 10 - x$ ... $(i)$ Let $A$ be the sum of double of the first and the square of the second: $A = 2x + y^2$ Substituting $y = 10 - x$ into the equation: $A = 2x + (10 - x)^2$ $A = 2x + 100 - 20x + x^2$ $A = x^2 - 18x + 100$ To find the minimum,differentiate $A$ with respect to $x$: $\frac{dA}{dx} = 2x - 18$ Set $\frac{dA}{dx} = 0$ for critical points: $2x - 18 = 0 \implies x = 9$ Check the second derivative: $\frac{d^2A}{dx^2} = 2$. Since $2 > 0$,the function has a minimum at $x = 9$. Substitute $x = 9$ into $(i)$: $y = 10 - 9 = 1$. Thus,the parts are $(9, 1)$.

Divide $10$ into two parts such that the sum of double of the first and the square of the second is minimum.

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