Find the minimum distance from the origin to the intersection of $xy=6$ with $7x+24z=0$.

Question

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.

Answer 1

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 $\sqrt{x^2+y^2+z^2}=\frac5{\sqrt2}$.