$l^{2} + m^{2} = 1$ then max value of $(l + m) =$ ?

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Solution

$l^{2} + m^{2} = 1$
$m^{2} = 1 - l^{2} \Rightarrow m = \sqrt{1 - l^{2}}$
Max value of $l + m$
$f (l) = l + \sqrt{1 - l^{2}}$
max value occurs at $f^{1} (l) = 0$
$1 + \frac{1}{2 \sqrt{1 - l^{2}}} (- 2 l) = 0$
$1 = \frac{l}{\sqrt{1 - l^{2}}}$
$1 - l^{2} = l^{2}$
${2 l}^{2} = 1$
$l = 1 / \sqrt{2}$ or $- 1 / \sqrt{2}$
$m = 1 / \sqrt{2}$ or $- 1 / \sqrt{2}$
max value of $l + m$ is $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

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