Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
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
add comment

1 Answer 1

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 $4xy$?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.