If x2 + y2 is fixed prove that x + y is maximized when x = y

Let $x$ and $y$ be positive real numbers such that $x^2 + y^2 = R$ is a constant. Prove that $x+y$ is maximized when $x = y$ .

Proof. From the equation $x^2 + y^2 = R$ we obtain,

$y = \sqrt{R - x^2}.$

Then, we want to find the maximum of the function

$f(x) = x + \sqrt{R - x^2} \quad \implies \quad f'(x) = 1 - \frac{x}{\sqrt{R - x^2}}.$

The derivative then has a zero when

$\frac{x}{\sqrt{R-x^2}} = 1 \quad \implies \quad x = \sqrt{\frac{R}{2}}.$

We then have

$\begin{align*} f'(x) &> 0 & \text{when} && x &< \sqrt{\frac{R}{2}} \\ f'(x) &< 0 & \text{when} && x &> \sqrt{\frac{R}{2}}. \end{align*}$

Therefore, $f(x)$ is increasing for $x < \sqrt{\frac{R}{2}}$ and decreasing for $x > \sqrt{\frac{R}{2}}$ ; hence, $f(x)$ has a maximum at $x = \sqrt{\frac{R}{2}}$ . By our equation for $y$ we then have

$y = \sqrt{R - x^2} = \sqrt{R - \frac{R}{2}} = \sqrt{\frac{R}{2}} = x.$

Hence, we indeed have $x+y$ maximal when $x = y. \qquad blacksquare$

Stumbling Robot

If x² + y² is fixed prove that x + y is maximized when x = y

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