# Minimum distance from the points of the function $\frac{1}{4xy}$ to the point $(0, 0, 0)$

I am trying to find the minimum distance from the points of the function $\large{\frac{1}{4xy}}$ to the point $(0, 0, 0)$.

This appears to be a problem of Lagrange in which my condition: $C(x,y,z) = z - \frac{1}{4xy} = 0$, and my function would be $f(x,y,z) = \sqrt{x^2+y^2+z^2}$ or if i'm correct, it would be the same as the minimum value I get from using thee function as $f(x,y,z) = x^2+y^2+z^2$.

If I do this, I would then have:

$$f(x,y,z,\lambda) = x^2 + y^2 + z^2 + \lambda(z-\frac{1}{4xy}) \\ = x^2 + y^2 + z^2 + \lambda z-\frac{1}{4xy} \lambda$$

Then findind the partial derivatives ($\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}, \frac{\partial f}{\partial \lambda}$) and solving for the values of $x, y, z, \lambda$ and the minimum value I get at the end would be my answer.

Would that be the correct solution?

• By substitution of the constraint we need to minimize $$\sqrt{x^2+y^2+\frac{1}{4x^2y^2}}$$ and we don't need Lagrange's method. In this case AM-GM is the most effective method. – user Jan 31 '18 at 13:23
• Sorry there was a mistake the function to minimize is of course $$\sqrt{x^2+y^2+\frac{1}{16x^2y^2}}$$ – user Jan 31 '18 at 14:28

## 2 Answers

Note that by substitution of the constraint we need to minimize

$$f(x,y)=\sqrt{x^2+y^2+z^2}=\sqrt{x^2+y^2+\frac{1}{16x^2y^2}}$$

and we don't need Lagrange's method.

In this case AM-GM is the most effective method, indeed

$$\frac{x^2+y^2+\frac{1}{16x^2y^2} }{3}\ge\sqrt{x^2\cdot y^2\cdot \frac{1}{16x^2y^2}}\iff x^2+y^2+\frac{1}{16x^2y^2}\ge3\sqrt{\frac1{16}}=\frac32\sqrt{\frac1{2}}$$

and equality holds when $$x^2=y^2=\frac{1}{16x^2y^2}\implies x=\pm\sqrt\frac1{16}=\pm\sqrt\frac14 \quad y=\pm x$$

Thus the minimum distance is

$$d_{min}=\sqrt{\frac32 \sqrt{\frac1{2}} } =\sqrt{\frac32}\sqrt{\frac1{2}}$$

attained for

$$P_1=\left(\sqrt\frac14,\sqrt\frac14,\frac14\sqrt16\right)\quad P_2=\left(\sqrt\frac14,-\sqrt\frac14,-\frac14\sqrt16\right)$$

$$P_3=\left( -\sqrt\frac14,\sqrt\frac14,-\frac14\sqrt16\right)\quad P_4=\left( -\sqrt\frac14,-\sqrt\frac14,\frac14\sqrt16\right)$$

As an alternative, since $f(x,x)>0$ its minimum coincide with the ninimum of

$$g(x,y)=f^2(x,y)=x^2+y^2+\frac{1}{16x^2y^2}$$

thus for the critical points we have

$$g_x=2 x - \frac1{8 x^3 y^2}=0$$ $$g_y=2 y - \frac1{8 x^2 y^3}=0$$

from which we obtain

$$x=\pm\sqrt\frac1{16}=\pm\sqrt\frac14 \quad y=\pm x$$

• I have not done minimizing before so this is a bit strange to be. – Omari Celestine Jan 31 '18 at 13:33
• @OmariCelestine You can proceed also with the usual method but you don't need Lagrange at all. And for simplicity you can minimize $f^2$. Note that in general AM-GM inequality are very powerful and important. – user Jan 31 '18 at 13:37
• What do you bean by the usual method? – Omari Celestine Jan 31 '18 at 16:25
• By partial derivatives, I add some detail on it! – user Jan 31 '18 at 16:31
• Exactly as I was proceeding? – Omari Celestine Jan 31 '18 at 16:46

Then you have the following equations:$$2x+\lambda\dfrac{1}{4x^2y}=0\\2y+\lambda\dfrac{1}{4xy^2}=0\\2z+\lambda=0\\z=\dfrac{1}{4xy}$$or$$\lambda=-8x^3y=-8xy^3=-2z=-\dfrac{1}{2xy}$$since $x,y\ne 0$ this leads to $x^2=y^2$ and $16x^4y^2=1$ therefore $16x^6=1$ and the answer is as following:$$x=k_1(\dfrac{1}{4})^{\dfrac{1}{3}}\\y=k_2(\dfrac{1}{4})^{\dfrac{1}{3}}\\z=\dfrac{1}{k_1k_2}(\dfrac{1}{4})^{\dfrac{1}{3}}$$where $$k_1=\pm 1\\k_2=\pm 1$$therefore we have 4 answers.