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(2). Consider the surface produced by the equation $xy^2z^3 = 2$ . Find the points in this surface closest to the origin?

So far here is what I have,

Let

$$ f(x,y,z) = x^2 + y^2 + z^2 $$

Where f is the square of the distance from the point $(x,y,z)$ to the origin. We want to minimize $f$, so our constraint is

$$ g(x,y,z) = xy^2z^3 - 2 $$

We must solve

$$\nabla f = \lambda \nabla g \quad \text{ and } \quad g = 0$$

By the former we must have

$$ \Bigg( \begin{matrix} 2x \\ 2y \\ 2z \\ \end{matrix} \Bigg) = \lambda \Bigg( \begin{matrix} y^2z^3 \\ 2xyz^3 \\ 3xy^2z^2 \\ \end{matrix} \Bigg) $$ Solving this system we get equations

$$ 2x = \lambda \text{ } y^2z^3 $$ $$ $$ $$ 2y = 2\lambda \text{ } xyz^3 $$ $$ $$ $$ 2z = 3\lambda \text{ } xy^2z^2$$ $$ $$


This is where I start to get stuck. From what I gathered from websites and youtube videos is that I would try to solve it using either cases, or by having the left side equal the same thing. I tried the latter and believe the answers were

$$ y^2 = 2x^2 \to y = \pm x\sqrt(2) \text{ and}$$ $$ 2z^2 = 3y^2 \to z = \pm y \sqrt{\frac{3}{2}} $$

I got these solutions by multiplying the first, second, and third equation by yz, xz, xy, respectively. Then having the first equation equal the second and then having the second equation = third equation. So it should look like this:

$$ 2xyz_1 = \lambda \text{ } y^3z^4 $$ $$ $$ $$ 2xyz_2 = 2\lambda \text{ } x^2yz^4 $$ $$ $$

$$ 2xyz_3 = 3\lambda \text{ } x^2y^3z^2$$ $$ $$

But from here I am stuck and I have no idea what to do or even if I am on the right path. Any help would be greatly appreciated.

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Hint

Note that $xyz\ne 0$ because if one is zero the other two must be zero. Thus we have that

$$\dfrac{2x}{y^2z^3}=\frac{2y}{2xyz^3}=\frac{2z}{3xy^2z^2}.$$ Simplifying the fractions we arrive at

$$\dfrac{2x}{y^2z^3}=\frac{1}{xz^3}=\frac{2}{3xy^2z}.$$ Now we cancel a $z$ in each denominator getting

$$\dfrac{2x}{y^2z^2}=\frac{1}{xz^2}=\frac{2}{3xy^2}.$$ From the first and the third we have $$6x^2y^2=2y^2z^2\implies z^2=3x^2.$$ From the second and the third $$3xy^2=2xz^2\implies 3y^2=2z^2\implies y^2=2x^2.$$ Thus

$$y=\pm\sqrt2 x,\quad z=\pm \sqrt 3 x.$$ Now use the constraint to get

$$\pm x\cdot 2x^2 \cdot 3\sqrt 3 x^2=2.$$

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  • $\begingroup$ Thank you @mfl for your answer. I'm having trouble with some of the values in your fractions. I understand the first simplify, but I'm having trouble understanding the second and third. How do you get xz^2 and in the third simplify you don't have a z at all. $\endgroup$ – Hawaiian Rolls Dec 5 '16 at 1:33
  • $\begingroup$ After simplifying the fractions we have a $z$ in all denominators. So we can cancel it. $\endgroup$ – mfl Dec 5 '16 at 6:50

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