# Solving a min max problem in 2 variables using Mathematica

## Homework Statement:

Solving min max problem in 2 variables using Mathematica

## Homework Equations:

(x + y)*Sin[x - y],

3 questions: how many min/max points in R^2 and how many "saddle points"(in 3d)?
Drawing the graph in 3d you see endless "mountains and valleys" which logic tells me there will also be infinite max min points in 2d regardless of where you slice the graph. apparently this is wrong and there is a finite max/min points in R^2/2D. Please note this problem does not have a domain.

f'x= (x + y) Cos[x - y] + Sin[x - y]==0
f'y =-(x + y) Cos[x - y] + Sin[x - y]==0

fx==fy
y =(-x)

f'x=sin2x==0 (infinite max min )

f'y=sin-2y==0 (infinite max min)

....what am i doing wrong?

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BvU
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Drawing the graph in 3d
Like this ? : apparently this is wrong
Where do you see the maxima ? (how are maxima defined ?)
Any definition of such a thing ? For a function of two variables ? Or do you mean 'for z = (x + y)*Sin[x - y] ' ?

What do you call 2D and what 3D ?

Note: You could also make a little coordinate transformation: u = x+y, v = x-y and study u * sin v

BvU
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fx==fy
y =(-x)

f'x=sin2x==0 (infinite max min )
f'y=sin-2y==0 (infinite max min)

....what am i doing wrong?
In the first place: What are you doing at all ?

Poor man's math:
maxima is where the rain accumulates when you hold the surface upside down

Mathematical criterion has to do with derivatives allright, but not in the way you treat it.

Ya I'm pretty sure ive done it correctly in terms of finding min and max points of a function with two variables.
Take the derivative with respect to each variable independently. and both ==0 you find the critical point. you can use various techniques to find out if its max or min.

Edit: Apparently the answer is 0, the function doesn't have any max or min points...

Last edited:
BvU
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function doesn't have any max or min points
That becomes clear from the picture. So what is it with the points that have $f_x=f_y=0$ ?

epenguin
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Gold Member
Zero in, zero in! You got x = -y.

Isn't that or is that a solution to part of the problem? If not why not?

And anyway doesn't it reduce the problem to one in one variable only? You can eliminate between your two equations for zero derivatives in more than one way. Last edited:
BvU