Absolute max/min

1. Nov 19, 2014

Panphobia

1. The problem statement, all variables and given/known data
Max/min of f(x, y) = 1/x + 1/y
Constraint 1/x^2 + 1/y^2 = 1

3. The attempt at a solution
This was a question on my midterm exam and I am just wondering what the right answer would be.
With lagrange multipliers, I got two candidate points (sqrt (2), sqrt (2)) and (-(sqrt (2),-sqrt (2)) then I did what the prof taught in class and I checked the end behaviour, I concluded that as either x or y go to infinity or negative infinity, the other variable goes to 1, now I wrote that there is no max or min, but I know I'm not right, what is the right answer and how would I get it?

2. Nov 19, 2014

Ray Vickson

Put $X = 1/x$ and $Y = 1/y$. Your problem becomes $\max / \min F(X,Y) = X+Y$, subject to the constraint $X^2 + Y^2 = 1$. The solutions are essentially obvious without calculus (but can be obtained via calculus, if you want): either $(X,Y) = (1/\sqrt{2},1\sqrt{2})$ or $(X,Y) = (-1/\sqrt{2},-1/\sqrt{2})$. (Just look at contour curves of $Z = X+Y$ and see how the relate to the circle $X^2+Y^2 = 1$.)

These are true maxima and minima, because they are nowhere near the disallowed points having $X = 0$ or $Y = 0$ (which would be at $\infty$ in the original variables $x,y$).

Last edited: Nov 20, 2014
3. Nov 19, 2014

Panphobia

Hmm, I still don't think I understand fully, I'll look at the problem.

4. Nov 19, 2014

Ray Vickson

The red line is the line of constant X+Y that is as far to the right as possible, while still touching the constraint curve. The blue line is the one as far as possible to the left.

5. Nov 19, 2014

Panphobia

On an exam I don't think with 50 minutes to do 6 long answer problems and 10 short answers we have time to plot the contour plot. Without looking at the contour plot, and just looking at the end behaviours how can you tell which ones are max/mins?

6. Nov 19, 2014

Ray Vickson

Try computing the numerical value of $f(x,y)$, to see which one is the largest. Anyway, you need not do lots of contour lines; just draw one or two, so you can understand what is happening. I will let you figure out the rest.

7. Nov 20, 2014

Staff: Mentor

I don't think this is the right way to approach things. Any time you can draw a quick sketch of things, you're going to be ahead of the game. In my experience as a college math teacher, students often think they will save time by not drawing a sketch, at the cost of getting a visual understanding of the problem, and consequently getting it wrong.

For this problem, making the realization that you can replace 1/x by X and 1/y by Y -- then all you need to do is plot a circle and a bunch of straight lines all of which have slope -1.

8. Nov 20, 2014

Panphobia

So this replacing 1/x with X and 1/y with Y, I should have come up by myself? Just a quick question, can you replace 1/x with X and 1/y with Y because as x or y approach infinity 1/x approachs 0 same with 1/y, so in X^2 + Y^2, X approaches 0, and Y approaches 1, and vice versa? The thing is, I didn't even know you could do that, is this a realization I am supposed to come to by myself? Or should I have been taught that?

9. Nov 20, 2014

Staff: Mentor

I think these fall under the heading of substitutions, which you probably have already been taught in your calculus course. These would be natural substitutions, given that the function and the constraint both involve 1/x and 1/y.

10. Nov 21, 2014

Panphobia

I will ask my professor today if we have learned substitutions, I don't remember learning them.

11. Nov 21, 2014

Panphobia

I just went to my professors office hours and he didn't even know you could do that. But he did give an explanation as to how the question is solved.

12. Nov 21, 2014

haruspex

I would think so.
You are right that in any such substitution you need to think about the ranges involved. Since x and y are by definition not infinite, X and Y cannot be zero, so you can discard such solutions. Correspondingly, the substitution does not allow you to consider x = 0 or y = 0, so you need to analyse those separately.

13. Nov 21, 2014

Staff: Mentor

Well, neither does the original function or the constraint, so this is easy.

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