Answer
The point $\left( {\frac{{48}}{{97}},\frac{{108}}{{97}}} \right)$ on the line $4x+9y=12$ is closest to the origin.
Work Step by Step
We are given the line $4x+9y=12$. Our task is to minimize the distance $d = \sqrt {{x^2} + {y^2}} $ subject to the constraint $g\left( {x,y} \right) = 4x + 9y - 12 = 0$.
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 $g\left( {x,y} \right) = 4x + 9y - 12 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2x,2y} \right) = \lambda \left( {4,9} \right)$
So, the Lagrange equations are
(1) ${\ \ \ \ }$ $2x = 4\lambda $, ${\ \ \ }$ $2y = 9\lambda $
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $\left( {0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$ and $y \ne 0$.
Thus, $\lambda = \frac{1}{2}x = \frac{2}{9}y$.
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2 we obtain $y = \frac{9}{4}x$.
Substituting it in the constraint gives
$4x + 9\left( {\frac{9}{4}x} \right) - 12 = 0$
$\frac{{97}}{4}x = 12$
So, $x = \frac{{48}}{{97}}$.
Using $y = \frac{9}{4}x$, we obtain the critical point: $\left( {\frac{{48}}{{97}},\frac{{108}}{{97}}} \right)$. This critical point corresponds to the minimum of $f$ subject to the constraint $g\left( {x,y} \right) = 4x + 9y - 12 = 0$.
Step 4. Calculate the critical values
We evaluate the minimum value $f\left( {\frac{{48}}{{97}},\frac{{108}}{{97}}} \right) = \frac{{144}}{{97}}$ (the minimum square of the distance).
Notice that there is no maximum value since there are points on the line that are arbitrarily far from the origin. Hence, the point $\left( {\frac{{48}}{{97}},\frac{{108}}{{97}}} \right)$ on the line $4x+9y=12$ is closest to the origin.
