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If $x+y=k, x>0, y>0$, then $x^2+y^2$ is minimum, if
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Verified Answer
The correct answer is:
$x=y$
$$
\begin{gathered}
x+y=k \\
\Rightarrow y=k-x, x>0, y>0
\end{gathered}
$$
Consider
$$
\begin{aligned}
& A=x^2+y^2=x^2+(k-x)^2=2 x^2-2 k x+k^2 \\
& \frac{d A}{d x}=4 x-2 k=0 \Rightarrow x=\frac{k}{2} \\
& \frac{d^2 A}{d x^2}=4>0
\end{aligned}
$$
$\therefore$ At $x=y=\frac{k}{2}$, we have
Minimum vale of $x^2+y^2$.
\begin{gathered}
x+y=k \\
\Rightarrow y=k-x, x>0, y>0
\end{gathered}
$$
Consider
$$
\begin{aligned}
& A=x^2+y^2=x^2+(k-x)^2=2 x^2-2 k x+k^2 \\
& \frac{d A}{d x}=4 x-2 k=0 \Rightarrow x=\frac{k}{2} \\
& \frac{d^2 A}{d x^2}=4>0
\end{aligned}
$$
$\therefore$ At $x=y=\frac{k}{2}$, we have
Minimum vale of $x^2+y^2$.
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