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What is the shortest distance from the surface $𝑥𝑦+15𝑥+𝑧^2=209$ to the origin?

My professor wasn't able to solve it with Lagrange multipliers when I asked him.

The easy way to solve it without Lagrange multipliers was solving for $z^2$ and then plugging it into the distance formula squared $D^2 = x^2 + y^2 + z^2 $ because when distance squared is minimum, distance is also minimum.

However, it seems a lot harder to solve through Lagrange multipliers.

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2 Answers 2

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We need to minimize $f(x,y,z)=x^2+y^2+z^2$ with the constraint $\phi(x,y,z)=𝑥𝑦+15𝑥+𝑧^2-209=0$ therefore we have

  • $2x=(y+15)\lambda$
  • $2y= x\lambda$
  • $2z= 2z\lambda \implies \lambda=1 \lor z=0$

for $z=0$ we obtain

  • $2x=(y+15)\lambda $
  • $2y= x\lambda \implies2y=x\frac{2x}{y+15}\implies 2(y+15)=x\frac{2x}{y+15}+30$
  • $𝑥𝑦+15𝑥-209=0 \implies x(y+15)=209$

and by $y+15=t$ we need to solve

  • $2t^2-30t=2x^2$
  • $tx=209$

that is

$$2t^4-30t^3-2\cdot 209^2=0$$

for $\lambda=1$ we obtain

  • $2x=(y+15)$
  • $2y= x$
  • $𝑥𝑦+15𝑥+z^2-209=0$
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  • $\begingroup$ I still get stuck on solving for the case of z=0. What would you do with that case. $\endgroup$
    – ebehr
    Nov 12, 2019 at 16:00
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    $\begingroup$ @ebehr Yes indeed it is not a simple case. We can obtain an equation of 4th degree. $\endgroup$
    – user
    Nov 12, 2019 at 16:32
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Solving with the Lagrange multipliers is also easier when you use minimization relative to squared distance rather than just distance -- the square roots do create some complications. (In fact, minimization allows you to replace any objective function with another that produces the same ordering of "heights".)

For Lagrangian $$ \mathcal{L} = x^2 + y^2 + z^2 +\lambda(xy+15x+z^2-209) \text{,} $$ we find \begin{align*} \frac{\partial\mathcal{L}}{\partial x} &= 2x+(15+y)\lambda \\ \frac{\partial\mathcal{L}}{\partial y} &= 2y + x \lambda \\ \frac{\partial\mathcal{L}}{\partial z} &= 2z(1+\lambda) \\ \frac{\partial\mathcal{L}}{\partial \lambda} &= 15 x + xy + z^2 - 209 \text{.} \end{align*} From the partial derivative with respect to $z$, $\mathcal{L}$ is stationary only if $z = 0$ or $\lambda = -1$. From the partial derivatives with respect to $x$ and $y$, we can solve for $x$ and $y$: if $\lambda^2 -4 \neq 0$, $x = \frac{30 \lambda}{\lambda^2 - 4}$ and $y = \frac{-15 \lambda^2}{\lambda^2 - 4}$.

Our first attempt at a solution is $$ 15 \frac{30 \lambda}{\lambda^2 - 4} + \frac{30 \lambda}{\lambda^2 - 4} \cdot \frac{-15 \lambda^2}{\lambda^2 - 4} + (0)^2 - 209 = 0 \text{,} $$ which, after much manipulation, gives that $\lambda$ is any of the four roots of $209\lambda^4 - 1672\lambda^2 + 1800 \lambda + 3344 $. Since neither $\lambda = 2$ nor $\lambda = -2$ are roots of this polynomial, we may use our rational expressions for $x$ and $y$. From this, $x$ is any root of $x^4+3135x-43681$, $y = \frac{x^3}{209}$, and $z = 0$. (There is a correspondence between choices of root for $\lambda$ and choices of root for $x$, but since $\lambda$ is not a coordinate of a solution, we do not consider it further.)

Our other attempt starts from $\lambda = -1$. Solving $\{ 2x - (15+y) = 0, 2y - x = 0\}$ gives $x = 10$, $y = 5$. Placing those into the constraint (the partial derivative with respect to $\lambda$), we obtain $z = \pm 3$

So we have five candidate minima:

  • $\lambda = -3.0160\dots$, $x = -17.753\dots$, $y = -26.772\dots$, $z = 0$, with squared distance $1031.9\dots$,
  • $\lambda = -1.0187\dots$, $x = 10.318\dots$, $y = 5.2558\dots$, $z = 0$, with squared distance $134.98\dots$,
  • $\lambda = 2.0174\dots \pm 1.0664 \mathrm{i}$, both of which we ignore because the resulting $x$, $y$, and squared distance are not real.
  • $\lambda = -1$, $x = 10$, $y = 5$, and $z = \pm 3$, with squared distance $134$.

As we can see, the minimum distance is $\sqrt{134}$, attained at $(x,y,z) = (10,5,\pm 3)$.

This can be done with the usual distance formula (not squared distance, as was done above). It is wise to replace $\sqrt{x^2 + y^2 + z^2}$ with a shorter name, perhaps "$\alpha$" during algebraic manipulations (while always remembering $\alpha$ contains $x$s, $y$s, and $z$s). Doing this, $$ \mathcal{L} = \sqrt{x^2 + y^2 + z^2} +\lambda(xy+15x+z^2-209) \text{,} $$ we find \begin{align*} \frac{\partial\mathcal{L}}{\partial x} &= \frac{x}{\alpha}+(15+y)\lambda \\ \frac{\partial\mathcal{L}}{\partial y} &= \frac{y}{\alpha} + x \lambda \\ \frac{\partial\mathcal{L}}{\partial z} &= z \left( \frac{1}{\alpha} + 2\lambda \right) \\ \frac{\partial\mathcal{L}}{\partial \lambda} &= 15 x + xy + z^2 - 209 \text{.} \end{align*}

Without "unwrapping" $\alpha$, the same procedure as above gives:

  • $\alpha + 1/\lambda = 0$, $x = 10$, $y = 5$, and $z = \pm 3$, or
  • $\alpha\lambda$ is any root of $209t^4 - 418t^2 + 225t + 209$, $x = \frac{209}{15}(1-(\alpha \lambda)^2)$, $y = \frac{209}{15} (\alpha \lambda)( (\alpha \lambda)^2 - 1)$, and $z = 0$.

This is the same set of five potential solutions (again, two roots aren't real, so we reject them quickly) as above.

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