Answer
The point $\left( {\frac{1}{{{2^{1/4}}}},\frac{{{2^{1/4}}}}{{2 - \sqrt 2 }}} \right)$ on the curve $y = x + {x^{ - 1}}$ is closest to the origin.
Work Step by Step
We have the curve $y = x + {x^{ - 1}}$.
Write $g\left( {x,y} \right) = x + {x^{ - 1}} - y = 0$.
Our task is to minimize the distance $d = \sqrt {{x^2} + {y^2}} $ subject to the constraint $g\left( {x,y} \right) = x + {x^{ - 1}} - y = 0$ in the first quadrant.
Since finding the minimum distance $d$ is the same as finding the minimum square of the distance ${d^2}$, our task is to minimize $f\left( {x,y} \right) = {x^2} + {y^2}$ subject to the constraint $g\left( {x,y} \right) = x + {x^{ - 1}} - y = 0$ for $x,y > 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1 of Section 15.8, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2x,2y} \right) = \lambda \left( {1 - \frac{1}{{{x^2}}}, - 1} \right)$
So, the Lagrange equations are
(1) ${\ \ \ \ }$ $2x = \lambda \left( {1 - \frac{1}{{{x^2}}}} \right)$, ${\ \ \ }$ $2y = - \lambda $
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Notice that the first equation of (1) does not allow $x=0$. Therefore, $\lambda \ne 0$. Thus, $x \ne 0$ and $y \ne 0$.
So, $\lambda = \frac{{2{x^3}}}{{{x^2} - 1}} = - 2y$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2 we obtain $y = - \frac{{{x^3}}}{{{x^2} - 1}}$.
Substituting it in the constraint gives
$x + {x^{ - 1}} + \frac{{{x^3}}}{{{x^2} - 1}} = 0$
$\frac{{{x^2} + 1}}{x} + \frac{{{x^3}}}{{{x^2} - 1}} = 0$
Multiplying both sides by $x\left( {{x^2} - 1} \right)$ gives
${x^4} - 1 + {x^4} = 0$
$2{x^4} - 1 = 0$
$\left( {\sqrt 2 {x^2} - 1} \right)\left( {\sqrt 2 {x^2} + 1} \right) = 0$
$\sqrt 2 {x^2} - 1 = 0$, ${\ \ \ \ }$ $\sqrt 2 {x^2} + 1 = 0$
So, the real solutions are $x = \pm \frac{1}{{{2^{1/4}}}}$.
Using $y = - \frac{{{x^3}}}{{{x^2} - 1}}$, we obtain the critical points: $\left( {\frac{1}{{{2^{1/4}}}},\frac{{{2^{1/4}}}}{{2 - \sqrt 2 }}} \right)$, $\left( { - \frac{1}{{{2^{1/4}}}},\frac{{{2^{1/4}}}}{{ - 2 + \sqrt 2 }}} \right)$.
Since the point is to be in the first quadrant, the critical point concerned is $\left( {\frac{1}{{{2^{1/4}}}},\frac{{{2^{1/4}}}}{{2 - \sqrt 2 }}} \right)$.
Step 4. Calculate the critical values
We evaluate $f$ at the critical point $\left( {\frac{1}{{{2^{1/4}}}},\frac{{{2^{1/4}}}}{{2 - \sqrt 2 }}} \right)$ and obtain $f\left( {\frac{1}{{{2^{1/4}}}},\frac{{{2^{1/4}}}}{{2 - \sqrt 2 }}} \right) = 2\left( {1 + \sqrt 2 } \right)$, which is the minimum square of the distance.
Notice that there is no maximum value since there are points on the curve that are arbitrarily far from the origin. Hence, the point $\left( {\frac{1}{{{2^{1/4}}}},\frac{{{2^{1/4}}}}{{2 - \sqrt 2 }}} \right)$ on the curve $y = x + {x^{ - 1}}$ is closest to the origin.
