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Need help in finding extreme values

  1. Jul 1, 2012 #1
    My teacher has written this problem.

    Find the extreme values of the function f(x,y) = x^4 + y^4 - 2(x - y)^2 ≤ 0.

    But what does the "≤ 0" mean? I know literally it means less than 0 but it seems like an unusual function.

    In addition, he gave us some hints which I don't understand at all. Here it is,

    from the partial derivatives,

    f_x = 4x^3 - 4(x-y) = 0

    f_y = 4y^3 + 4(x-y) = 0

    we get (x,y) = (0,0),(√2,-√2),(-√2,√2) for O,A,B.

    (I think A is is f_xx(x,y) and B is f_xy(x,y) = 2 , but I am not sure because he didn't say anything)

    At the origin O, = (0,0),

    f_xx(0,0)x^2 + z.f_xy(0,0)xy + f_yy(0,0)y^2 = -4x^2 + 8xy - 4y^2

    = -4(x^2 - 2xy +y^2)

    = -4(x-y)^2 ≤ 0

    So that's the hint that he gave to the class. Can you explain this because I don't understand anything from it. It's different than how I would usually find extreme values.
  2. jcsd
  3. Jul 1, 2012 #2


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    hi aruwin! :smile:

    (try using the X2 and X2 buttons just above the Reply box :wink:)
    i think you must have copied that wrong, i think the ≤ 0 is a comment the teacher made when telling you how to check that it's not a saddle point, and has nothing to do with the definition of f
    if you add fx + fy, you get 4x3 = -4y3,

    so x = -y, and you can easily prove that x = 0 √2 or -√2

    (and he's arbitrarily called those solutions O A and B)
    the formula fxx(0,0)x2 + 2.fxy(0,0)xy + fyyy(0,0)y2 is the formula for checking whether you have a maximum minimum or saddle point

    (there should be a proof of that somewhere in your notes)

    he's evaluated everything at x = y = 0 (ie, point O), eg fxx = 12x2 - 4 at x = 0, ie fxx|x=0 = -4 :wink:
  4. Jul 1, 2012 #3
    No, I didn't copy it wrong. I have asked him again what it means and he said that it's the region where z = f(x,y) ≤ 0.
    Without "f ≤ 0", f -> infinity as x,y -> infinity, thus no finite upper bound.
    And this part:
    At the origin O = (0,0),
    f_xx(0,0)x^2 + z.f_xy(0,0)xy + f_yy(0,0)y^2 = -4x^2 + 8xy - 4y^2
    = -4(x^2 - 2xy +y^2)
    = -4(x-y)^2 ≤ 0

    is a taylor expansion.

    This whole thing is confusing, I need help to understand this. I couldn't ask him further more because it was time :(
  5. Jul 1, 2012 #4


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    At the point where you get a maximum (even one in a multi-dimensional setting), all second derivatives should be negatives. It's a similar interpretation to the normal univariate case where you do a second-derivative test to check whether you have a local minimum or maximum.

    Before going further though, what kinds of concepts are you using in this course? Are you using Lagrange multipliers? Hessian matrices?

    Maybe you could give the readers a bit of pretext for this problem by mentioning the specific concepts that are tied to the lecture and the associated problem.
  6. Jul 1, 2012 #5
    The concepts in this lecture are about extreme values and applying taylor expansion.Does this give you any idea on how to solve this? And why is taylor expansion used in this problem?
  7. Jul 1, 2012 #6


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    A taylor expansion is basically an expansion of a function about a point. In short, it takes a function that has certain properties: differentiability being the key one, but also having a radius of convergence around the point of expansion.

    The idea is that you can represent any function by choosing some point and 'expanding' around that point so that you get a function in terms of f(x+a) instead of f(x).

    The reasons for doing this are varied, but one reason has to do with getting good approximations for when you need to get an approximation of a function at a point that is close to some other point (in this case the 'a' above), and you know the derivatives at the point a.

    Now in your expansion, we know that the first partial derivatives are 0 since they are extreme points, which means in the expansion itself, the df/dx and df/dy are 0 and we also know that f(0,0) = 0. This means that this expansion takes into account the first and second partial derivatives for the expansion of f(x,y) around the point (0,0).

    Does this help you?
  8. Jul 2, 2012 #7
    But why do we need to expand the function?And secondly,how to expand it?Is there any formula involved in that case?
  9. Jul 2, 2012 #8


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    There is a formula for the general multivariable case and it's basically summing all the derivatives at some centre of expansion while postmultiplying each term by a centred polynomial with respect to its coeffecient.

    This is probably a mouthful, so I'll just link you to the wiki site (which is also a mouthful).


    There is no one use for Taylor series: it's merely a way to represent a function (including multivariable function) in a different way, kind of like representing 2 + 2 = 1 + 3.

    The reason its useful is because it's used in many approximation applications where we typically centre around a point and ignore a majority of the terms (this is called a Taylor polynomial), and again it's not just used for one particular thing, but for mathematics results in general (usually again with approximation).

    The formula is located on the wiki, but I'm afraid you might get a little lost by reading the general definition.

    For the f(x,y) case, let's say we centre it around a and b (a is x co-ordinate, b is y co-ordinate). We then calculate all the partial derivatives at the (a,b) point. For the nth derivative we have N+1 terms, corresponding to all the partial derivatives that are possible for 2 variables (x,y). For other numbers of variables we basically look at the number of individual terms for the expansion of an n-nomial. So for five variables, we look at the number of individual terms for (x + y + z + w + u)^n and then each individual simplified term in terms of the multinomial expansion will correspond to a derivative.

    Lets stick to the case of only x and y. This means for every nth order term, we will have N partial derivatives. Lets look at the first couple:

    df/dx, df/dy (N = 1)
    d^2f/dx^2, d^2f/dxdy, d^2f/dy^2 (N=2)
    d^3f/dx^3, d^3f/dx^2dy, d^3f/dxdy^2, d^3f/dy^3 (N=3)

    and so on.

    These are evaluated at your centre of expansion (which in the above is (a,b) for 2D function f(x,y)). You then add the centred expansion corresponding to the derivative: again a few examples,

    df/dx*(x-a), df/dy(y-b), df^3/dx^2dy*(x-a)^2*(y-b)/3!, df^2/dy^2*(y-b)^2/2! (remember the division of n! for an n-th order derivative).

    Once you match the derivative with its proper expansion, you take all these terms and add them together and that gives the expansion around some centre (a,b) for the 2D case.

    Again for clarification look at the wiki or another source for definitions and more examples.
  10. Jul 2, 2012 #9
    Thanks for the explanation. OK, I think I know the purpose of taylor expansion now but what I don't get is how do I determine the maximum value and minum value of the function after expanding it?
  11. Jul 2, 2012 #10


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    Well if the function is continuous and differentiable, then the same idea is used for the multi-variable case as it is for the univariate (single-variable) case.

    What happens is that you find stationary points within some region and then test the nature of those points with information on the second derivative.

    Although this sounds simple, depending on your constraints and function, it can be pretty easy or pretty hard.

    For your example, you don't actually have hard constraints and because of this you can do first and second derivative tests to first find the stationary points (which you have done) and then find out what kind of point they are (minimum, maximum, point of inflection).

    If you need to introduce constraints, the first level used is what are known as Lagrange multipliers. From this you start to build up more complicated techniques which are the subject of a field known as Mathematical Optimization which considers the problems of whether global minima or global maxima even exist, and when they exist for more abstract problems.

    Are you or have you covered a topic known as Lagrange multipliers and its use for optimization?
  12. Jul 2, 2012 #11
    Yes, I need to introduce constraints in this problem and I have to find the global max and global minimum. But I haven't covered Lagrange multipliers yet. Note that sometimes the teacher gives us questions that are not fully taught yet.
  13. Jul 2, 2012 #12


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    Well you should probably look up Lagrange multipliers as this probably the most important foundational concept in multi-variable optimization.

    If it's not in your notes or your own resources, do a quick google search and access many of the notes on introductory multivariable calculus and optimization that talk about these.
  14. Jul 3, 2012 #13
    I am so sorry.I have made a mistake. I dont need to find the global maximum/minimum but the local maximum/minimum.
    OK, so after I have expanded the function at point (0,0),(√2,-√2) and (-√2,√2), how do I determine the local max/min values?
    At (0,0), using the taylor expansion formula :
    f_xx(0,0)x^2 + 2.f_xy(0,0)xy + f_yy(0,0)y^2 = -4x^2 + 8xy - 4y^2

    So, what do I do next?How do I determine the conditions for it to become local maximum and minimum?
    Also, I think another important point to take note is that in the beginning of the question,we are given f(x,y) ≤ 0 and therefore, the result of the expansion -4x^2 + 8xy - 4y^2 ≤ 0.

    Now I am stucked =( Help,please......
  15. Jul 3, 2012 #14


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    hi aruwin! :smile:

    ah, a local maximum/minimum!

    i think you're missing the overall picture …
    you have a roughly circular region on whose boundary f(x,y) = 0, and inside it f(x,y) is negative

    you've found, by differentiating once, that there can only be three extreme points, O A and B, and you know that A and B are outside the region, so that leaves only O

    so you differentiate again, and apply the formula, and you get a quadratic, in this case 4(x-y)2
    no, f(x,y) ≤ 0 is irrelevant, that quadratic would be negative anyway

    because it is negative, the "acceleration" in any direction (not just x and y) near O is negative, and so O is a maximum :smile:
  16. Jul 3, 2012 #15
    I am so sorry if I am not getting this as quickly as I should,so please be patient with me =)
    Ok, why are points A and B outside of the region?When I substitue the values of (√2,-√2)and (-√2,√2) into the original function f(x,y), the result is negative. So doesnt that mean they are inside the region?
  17. Jul 3, 2012 #16


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    sorry, yes, they are in the region …

    ok, so now you need to check A and B also! :smile:
  18. Jul 3, 2012 #17
    Oh no, now I realize another mistake that I made. There was a mistake in the taylor expansion. I forgot to divide by the factorial nth term which is 2!OK, now the result of the expansion should be : -2(x-y)^2 (I think it cant be expanded more because after the second partial differentiation, there's no more x and y left)

    Here,I understand that the quadratic will always be a negative because of the "-2" at the front.

    OK, and my teacher had just given the answer for point O (0,0). Here it is and please explain to me why is the answer so???I don't get it, I can't even imagine how the shape of the graph would turn out.

    Answer for point (0,0):
    For all values of (x,y) near (0,0),
    f(x,y)≥f(0,0) → At (0,0) f(x,y) has a local minimum
    f(x,y)≤f(0,0) → At (0,0) f(x,y) has a local maximum

    *And for the other two points, (√2,-√2)and (-√2,√2) I still haven't got the answers.
  19. Jul 3, 2012 #18


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    hi aruwin! :smile:
    that's not the answer for this f,

    that's the definition of "local minimum" and "local maximum" for any f …

    eg, if you can find a neighbourhood of (0,0) such that f(x,y) is always higher than f(0,0), then we say that there is a local minimum at (0,0)​

    the reason why the fxx etc fomula works is that you need to check the sign of fkk for any direction k (not just x and y)

    and fkk is, approximately, a linear combination of fxx fxy and fyy
  20. Jul 3, 2012 #19
    Oh, I thought he was giving the answer lol. So he was just giving the definition.
    OK, so how do I know if f(x,y) is greater or lesser than f(0,0)?? What I understand is that from the taylor expansion, we know that f(0,0) is always negative (or 0).In other words, f(x,y) is always negative(or 0) at (0,0). Right?
    And since at the beginnging of the problem, the constraint f(x,y) ≤ 0 is given, it is also negative or 0 but how to know that it's bigger or than f(0,0)? If I can understand this part, I can conclude that O has a local maximum or minimum value of 0. Although I am pretty sure that it has a maximum value at O because the rest of the values of x and y are below zero.
    Last edited: Jul 3, 2012
  21. Jul 3, 2012 #20


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    if z = f(x,y) is a map of the ground, then we know that if there's a cross-roads at O, with the roads going north-south and east-west,

    then from fx and fy, we know both roads are flat at O

    and from fxx and fyy, we also know that both roads have a maximum at O​

    but we don't know whether a road going from P = (-x,-y) to Q= (x,y) has a maximum at O

    so we use that quadratic formula to give us the test for that road, POQ

    if that test shows us that POQ always has a maximum, wherever P is, then obviously there is a local maximum at O, ie a little hilltop at O
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