The two positive numbers with sum t,and the s | vedclass

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Question

(C) Let the two positive numbers be $x$ and $y$. Given that $x + y = t$,so $y = t - x$. We want to minimize the sum of their squares,$S = x^2 + y^2$.

Vedclass · Accepted Answer

$\frac{t}{2}, \frac{t}{2}$

Vedclass · Answer

(C) Let the two positive numbers be $x$ and $y$. Given that $x + y = t$,so $y = t - x$. We want to minimize the sum of their squares,$S = x^2 + y^2$. Substituting $y = t - x$,we get $S(x) = x^2 + (t - x)^2 = x^2 + t^2 - 2tx + x^2 = 2x^2 - 2tx + t^2$. To find the minimum,we take the derivative with respect to $x$: $\frac{dS}{dx} = 4x - 2t$. Setting $\frac{dS}{dx} = 0$,we get $4x = 2t$,which implies $x = \frac{t}{2}$. Since $\frac{d^2S}{dx^2} = 4 > 0$,the function has a minimum at $x = \frac{t}{2}$. Then $y = t - \frac{t}{2} = \frac{t}{2}$. Thus,the two numbers are $\frac{t}{2}$ and $\frac{t}{2}$.

The two positive numbers with sum $t$,and the sum of their squares is minimum are

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