Find the minimum distance from the origin to the intersection of $xy=6$ with $7x+24z=0$.
Solution:
We should minimize $f(x,y,z)=x^2+y^2+z^2$ subject to the two conditions $g(x,y,z)=xy-6$ and $h(x,y,z)=7x+24z-0$. By the Lagranage multiplier method we , we have:$$\triangledown f=\lambda\triangledown g+ \mu\triangledown h \implies 2x \hat{i}+2y\hat{j}+2z\hat{k}=\lambda(7\hat{i}+24\hat{k})+\mu(y\hat{i}+x\hat{j})$$From here, we have $$\begin{align*}2x=7\lambda+\mu y, \;\;2y=\mu x,\;\;2z=24\lambda \end{align*}$$
Now, we should solve these equations with $xy=6$ and $7x+24z=0$. How should I proceed to solve for $x,y,z$ and $\mu,\lambda$. Thanks.
$\endgroup$ 12 Answers
$\begingroup$Number the equations:$$2x=7\lambda+\mu y\tag1$$$$2y=\mu x\tag2$$$$2z=24\lambda\tag3$$$$xy=6\tag4$$$$7x+24z=0\tag5$$We solve them as follows:
- From $(2)$ and $(4)$ we get $(6):12=\mu x^2$.
- From $(1)$ and $(4)$ we get $(7):2x=7\lambda+6\frac\mu x$.
- From $(7)$ and $(6)$ we get $(8):2x=7\lambda+\frac{72}{x^3}$.
- From $(8)$ and $(3)$ we get $(9):2x=\frac7{12}z+\frac{72}{x^3}$.
- From $(9)$ and $(5)$ we get $(10):2x=-\frac{49}{288}x+\frac{72}{x^3}$, or$$\frac{625}{288}x=\frac{72}{x^3}$$$$x^4=\frac{20736}{625}\implies x=\pm\frac{12}5$$From relations $(2)$ to $(5)$, we see that we will get symmetric solutions for the remaining variables upon changing the sign of $x$. The points where the intersection curve comes closest to the origin are$$(x,y,z)=\left(\pm\frac{12}5,\pm\frac52,\mp\frac7{10}\right)$$and the minimum distance is $\sqrt{x^2+y^2+z^2}=\frac5{\sqrt2}$.
As an alternative to Lagrange multipliers, the intersection of the two sets consists of the points $\left(x,\frac{6}{x},-\frac{7x}{24}\right), x\ne 0$, so we have to minimize$$f(x)=x^2+\frac{36}{x^2}+\frac{49x^2}{576}=\frac{625x^4+20736}{576x^2} \\ f'(x)=\frac{625x^4-20736}{288x^3}$$Setting $f'$ to $0$ and examining its sign, we find the minimum of $f$ is obtained when $x=\pm2.4$ and so the shortest distance is$$\sqrt{f(\pm 2.4)}=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt 2} $$
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