Answer
The point on the ellipse with the largest $x$-coordinate is $\left( {8, - 2} \right)$.
Work Step by Step
In this example, our task is to maximize the function $f\left( {x,y} \right) = x$ subject to the constraint $g\left( {x,y} \right) = {x^2} + 6{y^2} + 3xy - 40 = 0$.
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {1,0} \right) = \lambda \left( {2x + 3y,12y + 3x} \right)$
The Lagrange equations are
(1) ${\ \ \ \ }$ $\lambda \left( {2x + 3y} \right) = 1$, ${\ \ \ }$ $\lambda \left( {12y + 3x} \right) = 0$
Since $\left( {0,0} \right)$ does not satisfy the constraint, we may assume that $x \ne 0$ and $y \ne 0$.
The first equation of (1) implies that $\lambda \ne 0$. Thus, from the second equation of (1) we get
$12y + 3x = 0$
$y = - \frac{1}{4}x$
Substituting $y = - \frac{1}{4}x$ in the constraint gives
${x^2} + 6{\left( { - \frac{1}{4}x} \right)^2} + 3x\left( { - \frac{1}{4}x} \right) - 40 = 0$
${x^2} + \frac{3}{8}{x^2} - \frac{3}{4}{x^2} - 40 = 0$
$\frac{5}{8}{x^2} - 40 = 0$
So, $x = \pm 8$.
Using $y = - \frac{1}{4}x$ we obtain the critical points: $\left( {8, - 2} \right)$ and $\left( { - 8,2} \right)$.
Now, we calculate the extreme values:
$f\left( {8, - 2} \right) = 8$
$f\left( { - 8,2} \right) = - 8$
Thus, we conclude that the point on the ellipse with the largest $x$-coordinate is $\left( {8, - 2} \right)$.
