Lagrange Multipliers (Question)

**VkL** · March 6th 2010, 04:58 PM

Hello, I am trying to find the absolute maximum and/or minimum of

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

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

This is what I have so far,

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

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

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

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

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.

**Soroban** · March 6th 2010, 05:34 PM

Hello, VkL!

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

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

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

Set the partial derivative equal to zero and solve:

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

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

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

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

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

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

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

**matheagle** · March 6th 2010, 10:48 PM

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).

**VkL** · March 7th 2010, 03:24 AM

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!

**HallsofIvy** · March 7th 2010, 03:57 AM

Since the Lagrange multiplier method typically results in equations like $f_1(x,y,z)= \lambda g_1(x,y,z)$ , $f_2(x,y,z)= \lambda g_2(x,y,z)$ , etc. and the value of $\lambda$ itself is not relevant to the solution, I find it useful to divide one equation by another, immediately eliminating $\lambda$ .

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

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

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

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