# Maximizing the function

1. Nov 19, 2007

### rye

1. The problem statement, all variables and given/known data
max f(x, y) = 2xy + 2y − x^2 − 2y^2
I need to get the value of x and y and determine if it gives the global maximum by using the infinity approach only.
Boundary is Negative infinity and Positive infinity

2. Relevant equations
max f(x, y) = 2xy + 2y − x^2 − 2y^2

3. The attempt at a solution
I am aware of the first step:
Partial differentiate with respect to x ++++++ 2y − 2x = 0
Partial differentiate with respect to y ++++++ 2x+ 2− 4y = 0
Solving both the equations will give x = y = 1

From here, I do not know how to use f(±INFINITY,±INFINITY) to determine if x = y = 1 yields the global maximum.

Your help is greatly appreciated. I have an exam next week on this.

Thanks Rye.

2. Nov 19, 2007

### HallsofIvy

Staff Emeritus
If you are using partial derivatives, this is hardly "pre-calculus"! I am going to move to this to calculus and beyond.

I don't know what your teacher means by "the infinity approach" but you DON'T "use f(±INFINITY,±INFINITY)" such a thing does not exist. You can, however, look at LIMITS as x and y go to infinity. In particular, along the line y= x, the function becomes f(x, x) = 2x2 + 2x − x2 − 2x2= 2x- 2x2. What happens to that as x goes to infinity? I particularly suggested y= x because f(x,y) = 2y- (x-y)2.

Again, this may not be the "infinity approach" but letting u= x-y, v= x+ y and writing f as a function f u and v might be enlightening.

Last edited: Nov 19, 2007
3. Nov 19, 2007

### rye

Hi,

Thanks for the reply. What I meant was the functions end points are positive and negative infinity. In order to test if f(1,1) qualifies to be the max value, I have to test out f(+∞,+∞) and f(-∞,-∞).

I guess what your mean by as x and y approaches infinity makes sense with f(+∞,+∞) and f(-∞,-∞).

However, I do not know what is the step from here. Pls enlighten. Thanks!

4. Nov 19, 2007

### HallsofIvy

Staff Emeritus
NO, what I meant was that f(+$\infty$, +$\infty$) have no meaning. You may be using them as shorthand for limits but you should be clear about that. Actually, since this is a function of two variables, you need more than just the limits as x and y individually go to infinity. You need to consider the possibility of (x,y) going infinitely far from (0,0) along any path. Have you looked at my last suggestion? If you take u= x-y and v= x+ y, then 2y= v- u, 2x= u+v and 2xy= (v^2- 8y^2)/2 so f(x,y)= 2xy- (x-y)^2= v^2/2- u^2/2- u^2= v^2/2- 3u^2/2. What kind of figure is that?

5. Nov 19, 2007

### rye

Hi HallsofIvy,

I am totally confused. I do not know where did u= x-y and v= x+ y come from ?

I am stuck at the step where I have got x = y = 1. From here how can I determine that indeed f(1,1) gives the global maximum? I would be grateful if you could tell me step by step.. Plsss

Tks.

Rye

6. Nov 21, 2007

### rye

anyone can help pls??