# Lagrange Multipliers (Question)

• Mar 6th 2010, 04:58 PM
VkL
Lagrange Multipliers (Question)
Hello, I am trying to find the absolute maximum and/or minimum of

$\displaystyle f(x,y)=xy$ on the circle $\displaystyle x^2 + y^2=1$

I am stuck and would like some help if any1 can help me.

This is what I have so far,

$\displaystyle \nabla f(x,y)=\lambda (g(x,y))$

$\displaystyle \nabla f(x,y)=yi+xj$
$\displaystyle \nabla g(x,y)=\lambda (2xi+2yj)$

setting the components equal to each other:
$\displaystyle y=2x\lambda$
$\displaystyle x=2y\lambda$
constraint: $\displaystyle x^2 + y^2=1$

I am having trouble solving for $\displaystyle x, y, \lambda$

I have 3 equations (including constraint) and 3 unknowns, It should work out. Can anyone help me? This is an algebra question, I feel like a moron.
• Mar 6th 2010, 05:34 PM
Soroban
Hello, VkL!

Sorry, I don't understand the method you were taught.

Quote:

Find extreme values of: .$\displaystyle f(x,y)\:=\:xy\:\text{ on the circle }\:x^2 + y^2\:=\:1$

We have: .$\displaystyle F(x,y,\lambda) \;=\;xy + \lambda(x^2+y^2-1)$

Set the partial derivative equal to zero and solve:

. . $\displaystyle \begin{array}{ccccccc}F_x &=& y + 2x\lambda &=& 0 & [1] \\ \\[-3mm] F_y &=& x + 2y\lambda &=& 0 & [2] \\ \\[-3mm] F_{\lambda} &=& x^2+y^2-1 &=& 0 & [3] \end{array}$

Solve [1] for $\displaystyle \lambda\!:\;\;\lambda \:=\:-\frac{y}{2x}\;\;[4]$

Solve [2] for $\displaystyle \lambda\!:\;\;\lambda \:=\:-\frac{x}{2y}\;\;[5]$

Equate [4] and [5]: .$\displaystyle -\frac{y}{2x} \:=\:-\frac{x}{2y} \quad\Rightarrow\quad y^2\:=\:x^2\quad\Rightarrow\quad y \:=\:\pm x\;\;[6]$

Substitute into [3]: .$\displaystyle x^2+x^2-1 \:=\:0 \quad\Rightarrow\quad 2x^2\:=\:1 \quad\Rightarrow\quad x \:=\:\pm\frac{1}{\sqrt{2}}$

Substitute into [6]: .$\displaystyle y \:=\:\pm\frac{1}{\sqrt{2}}$

Therefore: .$\displaystyle \begin{Bmatrix}\text{Maximum:} & \left(\dfrac{1}{\sqrt{2}}\,,\:\dfrac{1}{\sqrt{2}}\ right) & \left(-\dfrac{1}{\sqrt{2}}\,,\:-\dfrac{1}{\sqrt{2}}\right) \\ \\[-3mm] \text{Minimum:} & \left(\dfrac{1}{\sqrt{2}}\,,\;-\dfrac{1}{\sqrt{2}}\right) & \left(-\dfrac{1}{\sqrt{2}}\,,\:\dfrac{1}{\sqrt{2}}\right) \end{Bmatrix}$

• Mar 6th 2010, 10:48 PM
matheagle
Both of these are the same.
Creating this F is the same as making the two gradients a multiple of each other.
The third equation is our constraint, g(x,y).
• Mar 7th 2010, 03:24 AM
VkL
Soroban, Thank you very much for clearly explaining step by step! It was solving for lambda I got stuck on. I don't know why. Once again, thank you very much!
• Mar 7th 2010, 03:57 AM
HallsofIvy
Since the Lagrange multiplier method typically results in equations like $\displaystyle f_1(x,y,z)= \lambda g_1(x,y,z)$, $\displaystyle f_2(x,y,z)= \lambda g_2(x,y,z)$, etc. and the value of $\displaystyle \lambda$ itself is not relevant to the solution, I find it useful to divide one equation by another, immediately eliminating $\displaystyle \lambda$.

In this case, setting $\displaystyle \nabla f(x,y,z)= \lambda\nabla g(x,y,z)$, which is the method I tend to use and is, of course, equivalent to Soroban's, we have $\displaystyle y= 2\lambda x$ and $\displaystyle x= 2\lambda y$.

Dividing the first by the second, $\displaystyle \frac{y}{x}= \frac{2\lambda x}{2\lambda y}= \frac{x}{y}$ which immediately gives $\displaystyle x^2= y^2$ and so $\displaystyle x= \pm y$.

Putting those into the constraint $\displaystyle x^2+ y^2= 1$ gives Soroban's solutions.