# LaGrange Multiplier Problem

• Feb 18th 2010, 06:19 PM
jaijay32
LaGrange Multiplier Problem
Hi. I have the equation x^2 + 4y^2, with the constraint xy=1. (with m being used to indicate theta)
Hence, F(x,y,m)= x^2 + 4y^2 + m(xy-1).

Derivatives:

Fx= 2x + my=0
Fy= 8y+ mx=0
Fm= xy-1=0.

What is the correct way to solve for y and x; I keep getting y= the sq. root of 1/2, but the book says this is wrong.

Many thanks
Jay
• Feb 18th 2010, 06:33 PM
Ted
Quote:

Originally Posted by jaijay32
Hi. I have the equation x^2 + 4y^2, with the constraint xy=1. (with m being used to indicate theta)
Hence, F(x,y,m)= x^2 + 4y^2 + m(xy-1).

Derivatives:

Fx= 2x + my=0
Fy= 8y+ mx=0
Fm= xy-1=0.

What is the correct way to solve for y and x; I keep getting y= the sq. root of 1/2, but the book says this is wrong.

Many thanks
Jay

$x^2 + 4y^2$ is not an equation.
• Feb 18th 2010, 06:44 PM
NOX Andrew
$f_{x} = 2x + my = 0 \implies m = -\frac{2x}{y}$

Substitute $m = -\frac{2x}{y}$ into $f_{y} = 8y + mx = 0$ to produce:

$f_{y} = 8y + (-\frac{2x}{y})x = 8y - \frac{2x^2}{y} = 0$.

Solving for y produces:

$y = \pm\frac{x}{2}$.

Substitute $y = \pm\frac{x}{2}$ into $f_{m} = xy - 1 = 0$ to produce:

$f_{m} = x(\pm\frac{x}{2}) - 1 = 0$.

Solving for x and disregarding the imaginary solutions produces:

$x = \pm\sqrt{2}$

Substituting $x = \pm\sqrt{2}$ into $y = \pm\frac{x}{2}$ produces:

$y = \pm\frac{\pm\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$
• Feb 19th 2010, 12:31 AM
elliotician
Quote:

Originally Posted by NOX Andrew
$f_{x} = 2x + my = 0 \implies m = -\frac{2x}{y}$

Substitute $m = -\frac{2x}{y}$ into $f_{y} = 8y + mx = 0$ to produce:

$f_{y} = 8y + (-\frac{2x}{y})x = 8y - \frac{2x^2}{y} = 0$.

Solving for y produces:

$y = \pm\frac{x}{2}$.

Substitute $y = \pm\frac{x}{2}$ into $f_{m} = xy - 1 = 0$ to produce:

$f_{m} = x(\pm\frac{x}{2}) - 1 = 0$.

Solving for x and disregarding the imaginary solutions produces:

$x = \pm\sqrt{2}$

Substituting $x = \pm\sqrt{2}$ into $y = \pm\frac{x}{2}$ produces:

$y = \pm\frac{\pm\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$

After substituting x you should get two sets of solutions:

$x=-\sqrt {2}, y=-1/2\,\sqrt {2}, m=-4$
$x=\sqrt {2}, y=1/2\,\sqrt {2}, m=-4$