1. ## Lagrange Multiplier Problem

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

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

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

Thus, yeilding my three equations with three unknowns:

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

$\displaystyle x=\lambda*(-1)$

$\displaystyle x^2-y+1=0$

Solving the equations, I get the following:

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

$\displaystyle x=-\lambda$

Plugging these into the third equation yeilds:

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

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

This is where I'm stuck, how can $\displaystyle \lambda^2$ (a positive number) plus $\displaystyle \frac{1}{2\lambda^2}$ (also a positive number) = -1 when added together. Can someone help me find my mistake so I can finish the problem? Thanks

2. ## Re: Lagrange Multiplier Problem

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

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

$\displaystyle x^2-y+1=0$
Right.

3. ## Re: Lagrange Multiplier Problem

The derivative of $\displaystyle \frac{x}{y}$ with respect to y, as FernandoRevilla suggests, is NOT "x" but $\displaystyle -\frac{y}{x^2}$. Write $\displaystyle \frac{y}{x}$ as $\displaystyle x^{-1}y$ and it should be easy to see why.

4. ## Re: Lagrange Multiplier Problem

Hello, dbakeg00!

I was taught a different method for Lagrange multipliers.

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

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

Set the partial derivatives equal to zero and solve.

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

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

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

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

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

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

5. ## 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.