1. ## working with min{f(x),f(y)}

How would you go about telling what the contour lines of a function such as min{x,y-1/x} has?

How, for example would you go about figuring out the kind of shape the graph would have?

I have graphed the function min{x, y-1/x} using wolfram but if am at a loss as to how I would figure out what it looks like without the aid of software.

A function such as min{ax,by} looks like a folded piece of paper along the line y=(b/a)x, but what, analytically speaking is the technique for gathering information on a general form of min{f(x),f(y)}

Also, what is this kind of problem called? If I googled min problems I generally get minimization problems.

2. ## Re: working with min{f(x),f(y)}

Hey kingsolomonsgrave.

You are probably better off to optimize both functions and find the situation when f(x) - f(y) < 0 and when f(x) - f(y) > 0.

More specifically this means finding when the two functions intersect along some boundary or surface and then looking at the derivatives at those points to see if the function is decreasing or increasing.

3. ## Re: working with min{f(x),f(y)}

First, the question in your title, to find max(f(x), f(y)) is not at all the question you are asking, to find the max of (x, y- 1/x). In the first you have the same function, f, but of the two different variables, x and y. In the second, you have two different functions, f(x)= x and g(x,y)= y- 1/x. To determine which is larger, you need to look at the function h(x,y)= f(x)- g(x,y)= x- (y- 1/x) and determine where it is positive and where negative. The crucial point is that those will be separated by the curve where it is 0 or where it is not defined.

It is 0 where f(x)- g(x,y)= x- y+ 1/x= 0. That, obviously, occurs on the curve y= x+ 1/x. The function is not defined where x= 0.

That divides the plane into three regions, one above the curve y= x+ 1/x for x> 0, one below the curve y= x+ 1/x for x< 0, and one between those two. The first contains the point (1, 3) because 1+ 1/1= 2< 3. The second contains the point (-1, -3) because -1+ 1/(-1)= -2< -3. The third contains the point (1, 1). Determine which of x and y- 1/x is larger in each of those regions by evaluating at one point in each region, perhaps the points I give.