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Find $max$ and $min$ of $M=2\sqrt{x+y} - \sqrt{x} - \sqrt {y}$ with $x + y = constant$

I tried to call $c = \sqrt{x+y} \Rightarrow M= 2c - (\sqrt x+\sqrt {y})\ge (2-\sqrt{2})c$ ?

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If you let $y=c-x$ then your formula for $M$ has only one variable, $x$, and usual calculus methods apply. Or, you can try Lagrange multipliers. Do you know those methods? It helps if we know what level the question comes from. – Gerry Myerson Feb 9 '13 at 5:14
$x \to \sqrt{x}$ is a concave function, which means that with $x+y$ fixed, $\sqrt{x} + \sqrt{y}$ is maximized when $x = y$, and minimized when they are as far apart as possible, in this case it means that one of $x,y$ is 0. – user27126 Feb 9 '13 at 5:29

Since $x+y$ is a constant, to maximize $2\sqrt{x+y}-\sqrt{x}-\sqrt{y}$, we need to minimize what we take away from $2\sqrt{x+y}$, so we need to minimize $\sqrt{x}+\sqrt{y}$.

Similarly, to minimize $2\sqrt{x+y}-\sqrt{x}-\sqrt{y}$, we need to maximize $\sqrt{x}+\sqrt{y}$.

So let us first minimize $\sqrt{x}+\sqrt{y}$. This is equivalent to minimizing its square, which is $x+y+2\sqrt{x}\sqrt{y}$. But $x+y$ is fixed. What is the minimum possible value of $2\sqrt{x}\sqrt{y}$?

Next we maximize $\sqrt{x}+\sqrt{y}$. Equivalently, we maximize its square $x+y+2\sqrt{x}\sqrt{y}$. So we want to maximize $4xy$.

Note that $(x+y)^2=(x-y)^2 +4xy$. So $4xy=(x+y)^2-(x-y)^2$. Given that $x+y$ is fixed, what does this say about the maximum value of $4xy$?

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