Find the 10 lowest values of the function

[TobeHode]

Homework Statement

Hi. I came across this problem last week and in order to solve it, I had to resort to excel. I believe there is a smarter way, but I can't seem to figure it out.

Given

$$f(x,y)=A(1+\frac{x^2}{4}+\frac{y^2}{9})$$

Where "A" is a constant, "x" and "y" are integer valued starting at 1 (1,2,3,...,n).

Determine the 10 lowest values of f(x,y).

Homework Equations

The Attempt at a Solution

To solve this, I used excel to quickly look at the first 10 or so values of f(x,y) for a specified x or y. For instance, I set x=1 and evaluated f(1,y) for y=1..10. I then set x=2 and repeated.

My first thought was to use Lagrange multipliers but I don't think that applies here because f(x,y) is not continuous. I am rusty on the method of Lagrange multipliers to minimize a function, so perhaps I am wrong.

Can anyone lend a hint as to a clever way to solve this?

 

[lanedance]

not A is increasing with both |x| and |y|, in fact the first 10 minimum values of x^2 + y^2 will be the first 10 minimums of A(x,y). And x^2 + y^2 is just the square of the distance from the origin...

it should be clear that (0,0) is the local (and global) minimum of A(x,y) considering it as a continuous function

then the next lowest values will be conatained in the positive quadrant with a given radius:
r = 1
(0,1), (0,1)
then r=sqrt(2)
(1,1)
and so on...

so i think the easiest way to do this is in essence just to count up the first 10

 

[TobeHode]

Thanks for the reply. I don't think I am following what you are saying. I shouldn't have used x and y. Let me write this problem in a different way:

$$f_{n_1,n_2} = \frac{n^{2}_{1}}{4}+\frac{n^{2}_{2}}{9}$$

where n_1 and n_2 are positive real integers

$$n_1:\left\{1, 2, 3,...,N}\right\}$$
$$n_2:\left\{1, 2, 3,...,N}\right\}$$

The first few lowest $$f_{n_1,n_2}s$$ are given by:


f_11
f_12
f_21
f_13
f_22
f_23
f_14

 

[Mark44]

What you have here is a function that is defined on a lattice of integer values. The simplest thing to do IMO is put the formula in a spreadsheet and note which values of x and y give you the smallest function values.

Assuming you have x values 1, 2, 3, ... going across the top of the spreadsheet, and y values 1, 2, 3, ... going down the left side, the smallest function values are going to be in the upper left corner. The smallest value is when x = 1, y = 1, and the next smallest is when x = 1, y = 2.

 

[lanedance]

yeah sorry i included 0 by mistake, otherwise everything i said applies