Lagrange multipliers and their system of equations

answered 2014-07-16 20:04:41 -0600

Justin
1009 ●2 ●20 ●38

updated 2014-07-16 20:58:32 -0600

Mark
166 ●1 ●5 ●12 https://sites.google.com/...

Given the system of equations:

$$ 2 - 2 \lambda x = 0 $$ $$ 4 - 2 \lambda x = 0 $$ $$ x^2 + y^2 = 20 $$

We can simply solve for $x$ and $y$ in terms of $\lambda$ from the first two equations and plug it in to the third equation.

Solving for $x$: $$ 2 - 2 \lambda x = 0 $$ $$ 2 \lambda x = 2 $$ $$ x = \frac{1}{\lambda} $$

Solving for $y$: $$ 4 - 2 \lambda y = 0 $$ $$ 2 \lambda y = 4 $$ $$ y = \frac{2}{\lambda} $$

Plugging in to the third equation: $$ \left(\frac{1}{\lambda}\right)^2 + \left(\frac{2}{\lambda}\right)^2 = 20 $$ $$ \frac{1}{\lambda^2} + \frac{4}{\lambda^2} = 20 $$ $$ \frac{5}{\lambda^2} = 20 $$ $$ \lambda^2 = \frac{5}{20} = \frac{1}{4} $$ $$ \lambda = \pm \frac{1}{2} $$

This means that there are $2$ cases to examine. When $\lambda = -\frac{1}{2}$: $$ x = \frac{1}{-\frac{1}{2}} = -2 $$ $$ y = \frac{2}{-\frac{1}{2}} = -4 $$

When $\lambda = \frac{1}{2}$: $$ x = \frac{1}{\frac{1}{2}} = 2 $$ $$ y = \frac{2}{\frac{1}{2}} = 4 $$

The two points that solve this system of equations are $(-2, -4)$ and $(2, 4)$.

Comment: Awesome!!