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I wish to solve the following problem:

Compute the minimum distance between the center to the curve $xy=4$.

But I don't know where to start from?

Niousha
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    Start with a picture. The answer will be obvious. If you do not trust the picture, minimize $\sqrt{x^2+y^2}$, given $xy=1$, or equivalently minimize $x^2+\frac{1}{x^2}$. You can use calculus, or other tools, such as AM/GM. – André Nicolas Dec 07 '13 at 19:06
  • Are you looking for a calculus solution, or a non-calculus solution? The comment above is fairly close to a solution. – André Nicolas Dec 07 '13 at 19:11
  • Calculus solution. I think this is an optimization problem, I was thinking of a more systematic approach to solve these kinds of problems. Your help is much appreciated. – Niousha Dec 07 '13 at 19:14
  • In my comments, there is a typo, I misread the $4$ as a $1$. Will type out an answer. – André Nicolas Dec 07 '13 at 19:17

3 Answers3

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We interpret the question as asking for the minimum distance from the origin to the curve $xy=4$. This is a rectangular hyperbola, with "centre" the origin.

The easiest way is to plot the curve, and note the symmetry. It is clear that the minimum is reached where $x=y$. Thus the nearest points to the origin are $(2,2)$ and $(-2,-2)$. The minimum distance is therefore $2\sqrt{2}$.

If we want to use calculus, note that the distance from $(x,y)$ to the origin is $\sqrt{x^2+y^2}$. We want to minimize $\sqrt{x^2+y^2}$ given that $xy=4$, that is, that $y=\frac{4}{x}$.

So we want to minimize $\sqrt{x^2+\frac{16}{x^2}}$, or equivalently to minimize its square. Let $$f(x)=x^2+\frac{16}{x^2}.$$ We want to minimize $f(x)$.

Note that $$f'(x)=2x-\frac{32}{x^3}.$$ Set this equal to $0$ and solve for $x$. We get $x=\pm 2$.

We need to verify that at $x=\pm 2$ we actually get a minimum. This can be shown in various ways. For example, for $x=2$, we can note that $f'(x)$ is negative in $(0,2)$, and positive in $(2,\infty)$. So in the interval $(0,2)$, the function $f(x)$ is decreasing, and thereafter it is increasing, so it must have reached a (local) minimum at $x=2$.

André Nicolas
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You want to minimise $D^2=x^2 + y^2$ subject to the condition $xy=4$.

$D^2=x^2 + \frac{16}{x^2}\ge2\sqrt{16}=8$ by AM-GM.

So the minimum distance is $\sqrt{8}$

user85798
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You need to minimize the function

$$d= \sqrt{x^2 +y^2} = \sqrt{x^2+ 16/x^2}.$$

Which we got by considering the distance from a point on the curve to the origin. Now, use the derivative test. See related problem.

Community
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Mhenni Benghorbal
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