Maximum of $x+y$ with constraint

Question

What is the maximum value of $x+y$ given that $x^2-4xy+4y^2+\sqrt{3x}+\sqrt{3y}-6=0$? $x,y$ are real numbers. Notice that it has terms $\sqrt{x}$ and $\sqrt{y}.$

If it is so easy, I wouldn't ask this question. The square roots of $x$ and $y$ makes the problem more difficult. — winterfly, Jun 06 '15 at 18:26
Do you need a solution in closed form, or would a numeric approximation suffice? — Rory Daulton, Jun 06 '15 at 18:47
I think, given $F(x,y)=0$ it suffice $\dfrac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}=-1$ to $x+y$ to have a maximum is that what @Mann have pointed. — Alexey Burdin, Jun 06 '15 at 19:23
@Mann: Agreed, but you would have obtained the same with Lagrange's method that you deemed overkill. I imagined that you had some really short and simple alternative method in mind. — Alex M., Jun 07 '15 at 10:02

score 0 · Answer 1 · answered Jun 07 '15 at 11:49

Uh well, after seeing this today. Here's a try. I actually didn't try to solve it yesterday.

$x^2-4xy+4y^2+\sqrt{3x}+\sqrt{3y}-6=0$ is the expression which can be rewritten as.

$(x-2y)^2=6-\sqrt{3}(\sqrt{x}+\sqrt{y})\geq 0$

Gives first condition as, $\sqrt{x}+\sqrt{y}\leq 2\sqrt{3} $

. Well this is for squareroot, not the expression we require but now we can just use a primitive tool, the AM-GM inequality to approach at,

$$\frac{\sqrt{x}+\sqrt{y}}{2}\geq (\sqrt{xy})^{1/2} $$

$$ (\sqrt{xy})^{1/2}\leq \frac{\sqrt{x}+\sqrt{y}}{2}\leq \frac{2\sqrt{3}}{2}$$

$$\sqrt{xy}\leq 3$$

now from our first condition squaring it gives.

$$x+y\leq 12-2\sqrt{xy} $$

now using above condition, gives.

$$x+y \leq 12-2\times 3 \leq 6$$

Probably not the most rigorous proof, and really there are some flaws on dependencies I'd say it's only a rough bound. But might be close to it. Pretty sure this is the cleanest answer you can get.

For $x=72-48\sqrt 2,y=36-24\sqrt 2$, $x+y=108-72\sqrt 2\approx 6.18$. — mathlove, Jun 07 '15 at 12:22

Maximum of $x+y$ with constraint

1 Answers1