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Math Help - Min/Max Questions

  1. #1
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    Min/Max Questions

    For a function f(x,y), the second derivative is equal to fxx*fyy - (fxy)^2

    My book says that if the second derivative is greater than 0, then a max/min exists at that point. If the second derivative is less than 0, then it's a saddle point.

    What if the second derivative is equal to 0?

    Also, how is it possible to test for the max/min of a function such as f(x,y,z)? I can find the critical points, but I don't know how to test if it's a min or a max.
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  2. #2
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    Hold on, if the first derivative is zero, the tangent to the function is a horizontal line (slope = zero) at that point or a horizontal plane, in the case of a two-parameter function. But consider the two dimensional case first. If the second derivative is zero, it means that the rate of change in the slope is zero. That occurs at an "inflection point" where the slope either stops increasing and starts decreasing, or vice versa. For example, the second derivative of the parabola f(x) = x^2 is just the number 2. It doesn't have any zeroes, so the slope continues to increase (become more positive as x increases from negative infinity to positive infinity. But the function f(x) = x^3 has a second derivative of 6x, and will be zero when x = 0. If you graph this and look at it you will see that as x increases from left to right, the slope is decreasing in the third quadrant until the point x=0, then increasing in the first quadrant. At the exact point where the curvature chances from "concave down" to "Concave up" there is an inflection point where the instantaneous rate of change of the slope is zero,

    In the three dimensional case, the analog to an inflection point is indeed a "saddle point".

    To test, evaluate the function for x plus or minus some small number, and the same for y. If the two values for x are less than the original value, the point is a maximum, ditto for y. If both x and y show a maximum, the point is the top of a "hill". If both a minimum, the bottom of a "hole." If one shows a maximum, the other a minimum, it is a saddle point.

    I don't think your textbook actually says what you quoted. It probably says that a point where the first derivative will be a maximum if the second derivative is negative at that same point. Is that helpful?
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  3. #3
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    Quote Originally Posted by KingMe View Post
    For a function f(x,y), the second derivative is equal to fxx*fyy - (fxy)^2

    My book says that if the second derivative is greater than 0, then a max/min exists at that point. If the second derivative is less than 0, then it's a saddle point.
    All of this is assuning that \nabla f= 0 at that point, right? However, that is NOT what your book says. If you want to go beyond partial derivatives to a more general of definition of "THE" derivative, you can use the matrix of second partials as "the derivative". That is done in some texts but never the determinant of that matrix which is what f_{xx}f_{yy}- f_{xy}^2 is. If that formula is positive and either f_{xx}or f_{yy} is negative, then the function has a maximum there. If that formula is positive and either f_{xx} or f_{yy} is positive, then the function has a minimum. If that formula is negative then the function has a saddlepoint.

    What if the second derivative is equal to 0?
    Then you can't tell. f(x, y)= x^4+ y^4, f(x,y)= -x^4- y^4, and f(x,y)= x^4- y^4 all have f_xx*f_yy- f_xy^2= 0 at (0, 0) but the first has a maximum there, the second a minimum, and the third a saddle point.

    Also, how is it possible to test for the max/min of a function such as f(x,y,z)? I can find the critical points, but I don't know how to test if it's a min or a max.
    It's a heck of a lot harder! It's really a question of the matrix of second derivatives which is
    \begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial x\partial y} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}
    for a function of two variables, x and y and
    \begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} & \frac{\partial^2 f}{\partial x\partial z}\\ \frac{\partial^2 f}{\partial x\partial y} & \frac{\partial^2 f}{\partial y^2}  & \frac{\partial^2 f}{\partial y\partial z} \\ \frac{\partial^2 f}{\partial x\partial z} & \frac{\partial^2 f}{\partial y\partial z} & \frac{\partial^2 f}{\partial z^2} \end{bmatrix}
    for a function of three variables, x, y, and z. At a fixed point, that is numerical valued matrix.

    What is important is its eigenvalues. If they are all positive, then the point is a minimum, if all negative, then the point is a maximum, if different signs then the point is a "saddle point".

    Notice that the formula you give, f_xx*f_xy- f_xy^2 is simply the determinant of the two by two matrix above. It can be shown that the determinant of a matrix is the product of all of its eigenvalues. If the determinant is positive that means the two eigenvalues have the same sign and so we have either a maximum or minimum.

    However, with three variables, the determinant is the product of three eigenvalues. If the product is positive that might mean they are all positive (so a minimum) or it might mean that one is positive and the other two negative.

    For anything more than two variables, you would have to actually determine the individual eigenvalues and not just their product.
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  4. #4
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    Alright, thanks for the information.

    I've never heard of eigenvalues before, so I'm assuming that's on the list of things that I'm not supposed to know yet.
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