Must find a function with specified max and min

[Boyle22]

This is a calc 3 problem dealing with partial derivatives in 3space:

Find a function (continuous, smooth) which has a local max at (1,1) and a local minimum at (-1,-1) and prove.

I know that for a function to have a max at a specified point (a,b):

Fx(a,b)=0
Fy(a,b)=0
D>0 where D=Fxx(a,b)Fyy(a,b)-[Fxy(a,b)]^2
D can not equal 0
Fxx(a,b)<0

For a min at specified point:

Fx(a,b)=0
Fy(a,b)=0
D>0 where D=Fxx(a,b)Fyy(a,b)-[Fxy(a,b)]^2
D can not equal 0
Fxx(a,b)>0So I've been spending the past few hours trying to figure out this problem and I'm not exactly sure how to go about doing it. I've just been making up functions by trial and error and haven't been getting anywhere really.

The one function that comes closest that I came up with was:

(x^2)(y^3)-(x^4)(y^6)+1/2(x^4)(y^6)

since x and y partials are both equal to 0 at point (1,1) and Fxx partial is negative... but when i plug in for D I get 0.

So what is the proper way of approaching this problem?

 

[christianjb]

My guess at a simple function would be

y=a*[(x-b)^3-(x-b)^2]+c

with a, b c to be determined by differentiation.

 

[Boyle22]

My guess at a simple function would be

y=a*[(x-b)^3-(x-b)^2]+c

with a, b c to be determined by differentiation.

I am a bit confused with how you came up with that equation. I need a function of two variables (x,y)... I think?

Something similar to this : f(x,y)=(x^2)(y^3)-(x^4)(y^6)+1/2(x^4)(y^6)

Thanks for your help.

 

[christianjb]

I am a bit confused with how you came up with that equation. I need a function of two variables (x,y)... I think?

A function of two variables defines a surface z=f(x,y). For every point in the x-y plane the function value gives its 'height'.

 

[arildno]

Well, it is a two-variable function you're after.

So, because you know that the second derivatives of quadratic functions equal some constant, they are clearly unsuitable, because the Hessian is needed to change its value at times.

Try therefore for a general cubic:
f(x,y)=ax^{3}+bx^{2}y+cxy^{2}+dy^{3}+ex^{2}+fxy+gy^{2}
Since you are essentially dealing with six restrictions, you should be able to determine paremeters a,b,c,d, e,f,g wirhout having to take into account linear terms as well.

 

[arildno]

Solve the four linear equations first, getting two free variables.

Then fiddle with those variables so that the Hessians at both points have the correct signs.

 

[christianjb]

OK- it sounds like I'm confused. A max at (1,1) could mean a max at y(1)=1, or it could mean a max at f(1,1).

 

[arildno]

The four equations are:
3a+2b+c+2e+f=0 (fx at (1,1))
b+2c+3d+f+2g=0 (fy at (1,1))
3a+2b+c-2e-f=0 (fx at (-1,-1))
b+2c+3d-f-2g=0 (fy at (-1,-1))

Probably, you can set e=f=g=0; and solve the resultantant system:
3a+2b+c=0
b+2c+3d=0

whereby c=-3a-2b, yielding d=2a+b

 

[Boyle22]

OK- it sounds like I'm confused. A max at (1,1) could mean a max at y(1)=1, or it could mean a max at f(1,1).

Sorry for not being clear. I need a max at f(1,1) and a min at f(-1,-1).

 

[Boyle22]

The four equations are:
3a+2b+c+2e+f=0 (fx at (1,1))
b+2c+3d+f+2g=0 (fy at (1,1))
3a+2b+c-2e-f=0 (fx at (-1,-1))
b+2c+3d-f-2g=0 (fy at (-1,-1))

Probably, you can set e=f=g=0; and solve the resultantant system:
3a+2b+c=0
b+2c+3d=0

whereby c=-3a-2b, yielding d=2a+b

Ok so the function has to be a cubic? Forgive me for being slow, I'm having a hard time understanding what's going on lol.

 

[arildno]

Ok so the function has to be a cubic? Forgive me for being slow, I'm having a hard time understanding what's going on lol.

Of course it doesn't have to be cubic!

There are zillions of functions satisfying your criteria.

However, out of CONVENIENCE, instead of looking after exotic functions that might satisfy your criteria, I choose to look for a boring polynomial because it is easy to deal with.

Understand the difference?

 

[Boyle22]

Of course it doesn't have to be cubic!

There are zillions of functions satisfying your criteria.

However, out of CONVENIENCE, instead of looking after exotic functions that might satisfy your criteria, I choose to look for a boring polynomial because it is easy to deal with.

Understand the difference?

Yes I understand. After I solve for all the coefficients a,b,c,d,e,f,g(still working on it) isn't there other criteria that must be met? Since you just set all the partials to 0 at the points of interest that just tells you they are critical points. From there they could be max's, mins or saddle points. Right? I'm still working it out now, but thank you for your help.

 

[arildno]

Yes I understand. After I solve for all the coefficients a,b,c,d,e,f,g(still working on it) isn't there other criteria that must be met? Since you just set all the partials to 0 at the points of interest that just tells you they are critical points.

Correct!

From there they could be max's, mins or saddle points. Right?

Right!
But tackle the inequalities at the end.
I am quite sure you won't need e,f,g to be any other numbers than zero.
The reason for this is that even without them, you'll get two free parameters out of the partials equations. That should give you enough fiddling space to satisfy the Hessian demands.

I'm still working it out now, but thank you for your help.

Good luck to you!



The rest of your criteria is that your function should be smooth&continuous, which is automatically satisfied by all polynomials.

 

[Boyle22]

Correct!

Right!
But tackle the inequalities at the end.
I am quite sure you won't need e,f,g to be any other numbers than zero.
The reason for this is that even without them, you'll get two free parameters out of the partials equations. That should give you enough fiddling space to satisfy the Hessian demands.


Good luck to you!



The rest of your criteria is that your function should be smooth&continuous, which is automatically satisfied by all polynomials.

I can't figure out the coefficients. :( I know this is just algebra and I should know it, but how do you solve for the 4 unknowns(a,b,c,d) with only the 2 equations(Fx,Fy)?

 

[Boyle22]

I think I'm heading down the right track with the Cubic function thanks to arildno but I can't figure out the coefficients. Anyone else out there that could help?