# Lagrange Multiplier Problem

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• Sep 26th 2011, 10:38 AM
dbakeg00
Lagrange Multiplier Problem
Solve the following problem using Lagrange multipliers. 1.) Find the critical points of f(x,y)= $\frac{x}{y}$ on the curve $y=x^2+1$.

$\nabla{f}=<\frac{1}{y},x>$

$\nabla{g}=<2x,-1>$

Thus, yeilding my three equations with three unknowns:

$\frac{1}{y}=\lambda*2x$

$x=\lambda*(-1)$

$x^2-y+1=0$

Solving the equations, I get the following:

$y=\frac{-1}{2\lambda^2}$

$x=-\lambda$

Plugging these into the third equation yeilds:

$(-\lambda)^2-\frac{-1}{2\lambda^2}=-1$

$\lambda^2+\frac{1}{2\lambda^2}=-1$

This is where I'm stuck, how can $\lambda^2$ (a positive number) plus $\frac{1}{2\lambda^2}$ (also a positive number) = -1 when added together. Can someone help me find my mistake so I can finish the problem? Thanks
• Sep 26th 2011, 11:51 AM
FernandoRevilla
Re: Lagrange Multiplier Problem
Quote:

Originally Posted by dbakeg00
$\frac{1}{y}=\lambda*2x$

Right.

Quote:

$x=\lambda*(-1)$
It should be $x/y^2=\lambda$

Quote:

$x^2-y+1=0$
Right.
• Sep 26th 2011, 12:38 PM
HallsofIvy
Re: Lagrange Multiplier Problem
The derivative of $\frac{x}{y}$ with respect to y, as FernandoRevilla suggests, is NOT "x" but $-\frac{y}{x^2}$. Write $\frac{y}{x}$ as $x^{-1}y$ and it should be easy to see why.
• Sep 26th 2011, 02:45 PM
Soroban
Re: Lagrange Multiplier Problem
Hello, dbakeg00!

I was taught a different method for Lagrange multipliers.

Quote:

Solve, using Lagrange multipliers.
Find the critical points of: . $f(x,y)\:=\:\frac{x}{y}$ on the curve $y\,=\,x^2+1$

$\text{We have: }\:g(x,y,\lambda) \;=\;\frac{x}{y} + \lambda(x^2+1-y)$

Set the partial derivatives equal to zero and solve.

. . $\begin{array}{ccccccc}\dfrac{\partial g}{\partial x} &=& \dfrac{1}{y} + 2x\lambda &=& 0 & [1] \\ \\[-4mm] \dfrac{\partial g}{\partial y} &=& \text{-}\dfrac{x}{y^2} - \lambda &=& 0 & [2] \\ \\[-3mm] \dfrac{\partial g}{\partial\lambda} &=& x^2+1-y &=& 0 & [3] \end{array}$

From [2]: . $\text{-}\frac{x}{y^2} - \lambda \:=\:0 \quad\Rightarrow\quad \lambda \,=\,\text{-}\frac{x}{y^2}$

Substitute into [1]: . $\frac{1}{y} + 2x\left(\text{-}\frac{x}{y^2}\right) \:=\:0$
. . $\frac{y-2x^2}{y^2} \:=\:0 \quad\Rightarrow\quad y-2x^2 \:=\:0 \quad\Rightarrow\quad y \:=\:2x^2\;\;[4]$

Substitute into [3]: . $x^2 + 1 - 2x^2 \:=\:0$
. . $x^2\:=\:1 \quad\Rightarrow\quad x \:=\:\pm1$

Sustitute into [4]: . $y \:=\:2(\pm1)^2 \quad\Rightarrow\quad y \:=\:2$

The critical points are: . $\left(1,2,\tfrac{1}{2}\right),\;\left(\text{-}1,2,\text{-}\tfrac{1}{2}\right)$

• Sep 27th 2011, 04:25 AM
dbakeg00
Re: Lagrange Multiplier Problem
Thanks Fernando & Halls...that was a dumb mistake which I should have caught. Soroban, I'll have to study your way for a bit and see how I like it. Thanks again to all for the help.