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The problem is as follows:

$$\min_{x,y} x + y$$

subject to

$$xy = 0.25$$

My attempt:

I used the method of Lagrange multipliers here setting

$f(x,y) = x + y$ and $g(x,y) = xy - 0.25 = 0$

So we have $\nabla f = \lambda \nabla f$

$\implies 1 = \lambda y$, and $1 = \lambda x$

$\implies x = y$ and from the constraint we get:

$x^2 = 0.25 \implies x = \pm \sqrt{0.25} = y$

Clearly we choose $x = y = - \sqrt{0.25}$ to minimize the function.

However, I checked this minimization problem with Wolfram Alpha and it says no global minima where found. Why is this and where did I go wrong? Thanks!

patrickh
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    Let $x\to-\infty$ and $y=0.25/x$ ; then $x+y\to-\infty$ while $xy=0.25$. You've found a local minimum. – Nicolas Mar 30 '16 at 08:51
  • @Nicolas so can it be said that this problem has no solution? – patrickh Mar 30 '16 at 08:54
  • Are you assuming $x,y\gt0$? – robjohn Mar 30 '16 at 08:55
  • @robjohn the question does not explicitly state that, but if it were the case what would the solution be? – patrickh Mar 30 '16 at 08:57
  • @patrickh If you're looking for a global infimum, then it is $-\infty$ ; if you're looking for a local minimum, then it is what you've found, so all depends on the context. Now, as robjohn has mentioned, it could be that there is(are) another constraint(s) on $x$ and $y$ to avoid any problem. – Nicolas Mar 30 '16 at 08:57
  • @Nicolas if we are assuming x,y > 0 would there be any solution? – patrickh Mar 30 '16 at 09:03
  • @patrickh Well, if $x,y>0$, then the Lagrange's multipliers theorem gives us the local minimum $x=y=\sqrt{0.25}=1/2$, so that $x+y\geq1$. Now you must check if it is a global minimum : to do this, consider the infimum of $x+y$ on the compact set ${(x,y)\in\mathbb{R}^2\mid x,y\geq0,x,y\leq R}$ for a fixed $R>0$. Use a continuity argument to show that $x+y$ reaches its infimum (so it is a minimum), and it must be the one found by the Lagrange's multipliers theorem. Now let $R\to+\infty$ and see if the minimum can become smaller. – Nicolas Mar 30 '16 at 09:11

2 Answers2

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The points where $xy=\frac14$ are shown in the blue hyperbolas below.

enter image description here

The red and yellow lines are the points where $x+y=1$ and $x+y=-1$.

The line $x+y=1$ is the furthest to the lower left that intersects the branch of the hyperbola in quadrant $1$. So if we restrict our attention to $x,y\gt0$, we get $x+y\ge1$. Thus, there is a minimum.

If we also consider $x,y\lt0$, then the line $x+y=-1$ is the furthest to the upper right that intersects the branch of the hyperbola in quadrant $3$. For the points in quadrant $3$, we have $x+y\le-1$. In this case, there is no minimum.

Note that if $x,y\gt0$, then $$ x+y-2\sqrt{xy}=\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\tag{1} $$ Since $xy=\frac14$, $(1)$ implies $$ x+y\ge1\tag{2} $$

robjohn
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As pointed out in the Comments under the question, if $x,y$ are allowed to be negative, we can get $x+y$ as small as we wish ($x=-N,y=-1/4N$). If we require $x,y$ to be positive, then AM/GM gives $\frac{x+y}{2}\ge\sqrt{xy}=\frac{1}{2}$, so $x+y\ge 1$ with equality when $x=y=\frac{1}{2}$.

almagest
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