# Lagrange multipliers to find max/min

1. May 4, 2014

### Feodalherren

1. The problem statement, all variables and given/known data

Use Lagrange Multipliers to find the minimum value of $f(x,y)=x^{2}+(y-1)^{2}$ that lie on the hyperbola $x^{2}-y^{2}=1$. Draw a picture to verify your final answer.

2. Relevant equations

$\nabla f=\lambda \nabla C$

3. The attempt at a solution
So I can find the critical points and plug them into f and get 3 on both. I can draw a sketch but it doesn't really tell me anything. What exactly am I looking for? I make a contour plot but I have no idea if it's a local max/min.. if it's a global max or min. Basically I can get the numbers but they aren't telling me anything.

2. May 4, 2014

### Ray Vickson

Show you work; I certainly get something different from you.

3. May 4, 2014

### LCKurtz

Show us your work for the critical points. And show us your contour plot on top of the graph of the hyperbola. It is important for understanding whether you have max, min, or neither.

4. May 4, 2014

### Feodalherren

I did essentially this - except for the last part which I don't understand.

5. May 4, 2014

### Ray Vickson

So, you want to minimize $x^2 + (y-2)^2$, NOT $x^2 + (y-1)^2$? How can we help if you give us the wrong problem?

6. May 4, 2014

7. May 4, 2014

### LCKurtz

I guess you mean what you don't get is interpreting what the contours are telling you. Suppose the xy plane is a thin metal plate and that function $x^2 + (y-2)^2$ represents the temperature at $(x,y)$. Your two points lie on the circle where the temperature is $3$. What is the temperature on the bigger circles? Can you see from that picture that the temperature on the hyperbola is never less than three? Is there a maximum temperature on the hyperbola? Note that there are temperatures less than $3$ but they aren't on the hyperbola. Do you get it now?

8. May 4, 2014

### Feodalherren

The temperature should be increasing as the circles get larger. Okay so if 3 is the shortest possible radius to the hyperbola I suppose I can see that it would be the lowest possible temperature and therefore a global min.
Correct?

9. May 4, 2014

### LCKurtz

"Should" be?? Is it or isn't it? The level curves are $x^2+(y-2)^2 = C$. Does $C$ get larger as the radii do?

$3$ is not the radius of the smallest circle that touches the hyperbola. It represents the temperature on that circle. You can tell the temperature at any point on the hyperbola by checking which circle passes through that point.

10. May 4, 2014

### Feodalherren

Okay now I'm confused again. It looks to me from your equation that C is the radius... Well the radius squared. Then by logic, yes, C gets larger as the radius gets larger.

Same thing here - they ALL look like global minimums. In fact, I can't even imagine what a maximum would look like unless the graph cuts off at some point. I also don't understand how he can say that x=0 is a solution in this one. It seems to me from the equation
2xy=2xλ that if x=0 then y=0 which doesn't obey the constraint.

11. May 4, 2014

### LCKurtz

If you want to discuss another problem please start a new thread. All I want to know right now is whether you understand the current problem. You never answered my question as to whether there is a maximum.

12. May 4, 2014

### Feodalherren

I just posted it as another example of not understanding the contour maps.
So in the problem we are discussing - no it doesn't seem like there is a maximum. But that's part of my confusion. It doesn't seem like there will EVER be a maximum in any graph unless the graph stops at some point.

13. May 4, 2014

### LCKurtz

That all depends on your function $f(x,y)$. Can you imagine if it had sines and cosines there might be relative maximums and minimums all over the place?

14. May 4, 2014

### Feodalherren

I'm not sure. Intuitively yes I can because they oscillate but I don't understand why or how I would see it.

15. May 5, 2014

### Ray Vickson

The points $(\sqrt{3/2},\sqrt{3/2})$ and $(-\sqrt{3/2},-\sqrt{3/2})$ in the first and third quadrants are maxima, while the points $(-\sqrt{3/2},\sqrt{3/2})$ and $(\sqrt{3/2},-\sqrt{3/2})$ in the second and fourth quadrants are minima of $f(x,y) = xy$. This obvious from the contour plots, if they are done correctly and carefully.

16. May 5, 2014

### Feodalherren

I beg to differ, it's not obvious at all :). I went to office hours today and I have a much better feel for it now. This problem is still giving me a hard time though.

How can he use the x=0 points when f isn't defined at x=0? Shouldn't they fall out of the solutions?

For the contour plots f becomes y = 1/x^2
Not defined for x=0.