Relative Extrema of Functions

(1) Definition (Relative Extrema of Functions) The function is said to have a relative maximum at if for all in an open disk containing and is a relative minimum if for all in an open disk containing Collectively, relative maxima and relative minima are called relative extrema.

(2) Proposition (Relative Extrema of Functions) If has a relative extremum at and partial derivatives and both exist at then

(3) Definition (Critical Point) A critical point of a function defined on an open set is a point in where either one of the following is true: (i) or (ii) at least one of or does not exist at A critical point is called a saddle point of if every open disk centered at contains points in the domain of that satisfy as well as points in the domain of that satisfy

(4) Example (Critical Point) Find the critical points for the given functions.

(a) Let Then



These partial derivatives are equal to 0 when and so the only critical point is By completing the square we find that



Since and we have for all values of and Therefore, is a local minimum, and in fact it is the absolute minimum of This can be confirmed geometrically from the graph of , which is the elliptic paraboloid with vertex as shown.



(b) Find the extreme values of

    Solution. Since and the only critical point is Notice that for points on the -axis we have so (if ) However for points on the -axis we have so (if ) Thus every disk with center contains points where takes positive values as well as points where takes negative values. Therefore cannot be an extreme value for so has no extreme values. This example illustrates the fact that a function need not have a maximum or minimum value at a critical point. The graph of is the hyperbolic paraboloid which has a horizontal tangent plane at the origin. You can see that is a maximum in the direction of the but not in the direction of the -axis. Near the origin the graph has the shape of a saddle.

(5) Proposition (Second Partials Test) Let have a critical point at and assume that has continuous second order partial derivatives in a disk centered at The discriminant of is the expression Then

    (i)  a relative maximum occurs at if and (or equivalently, and

    (ii)  a relative minimum occurs at if and (or equivalently, and

    (iii)  a saddle point occurs at if

If then the test is inconclusive.

(6) Example (Second Partials Test) Find the relative extrema of the following functions.

(a) Find the local extrema of



    Solution. We first locate the critical points:



Setting these partial derivatives to 0, we obtain the equations  



To solve these equations we substitute from the first equation into the second one. This gives



So there are three real roots: The three critical points are and Next we calculate the second partial derivatives and





Since it follows that the origin is a saddle point; that is, has no local extremum at Since and we see that is a local minimum. Similarly, we have and so is also a local minimum.    



(b) Find the shortest distance from the point to the plane

    Solution. The distance from any point to the point is



but if lies on the plane, then and so we have



We can minimize by minimizing the simpler expression



By solving the equations





we find that the only critical point is Since and we have and so has a local minimum at Intuitively, we can see that this local minimum is actually an absolute minimum because there must be a point on the given plane that is closest to If and then



Thus the shortest distance is

(c) A rectangular box without a lid is to be made from 12 of cardboard. Find the maximum volume of such a box.

    Solution. Let the length, width, and height of the box (in meters) be and Then the volume of the box is We can express as a function of just two variables by using the fact that the surface area of the sides and the bottom of the box is



Solving these equation for we get so the expression for becomes



We can compute the partial derivatives:



If is a maximum, then but or gives so we must solve the equations



These imply that and so . (Note that and must both be positive in this example.) If we put in either equation we get which gives and From the physical nature of this example there must be an absolute maximum volume that has to occur at a critical point of so it must be when and Then so the maximum volume of the box is

(7) Definition (Absolute Extrema) The function is said to have an absolute maximum at if for all in the domain of . Similarly, has an absolute minimum at if for all in Collectively, absolute maxima and minima are called absolute extrema.

(8) Proposition (Absolute Extrema) A function of two variables attains both an absolute maximum and an absolute minimum on any closed bounded set where it is continuous.

(9) Example (Absolute Extrema) Find the absolute maximum and minimum values of the given functions over the given region.

(a)   over the rectangle  



    Solution. Since is a polynomial it is continuous on the closed bounded rectangle   therefore has both absolute maximum and minimum values. We first find the critical points by solving the system   



The only critical point is and the value of there is We look at the values of on the boundary of , which consists of four line segments as shown.



On we have and for This is an increasing function of so its minimum value is and its maximum value is On we have and on This is a decreasing function of so its maximum value is and its minimum value is On we have and on By observing that we see that the minimum value of this function is and the maximum value is Finally, on we have and on with maximum value and minimum value Thus on the boundary, the minimum value of is 0 and the maximum is 9.  We compare these values with the value at the critical point and conclude that the absolute maximum value of on is and the absolute minimum value is



(b) over the rectangle



    Solution. We compute and and set and find that is the only critical point in the interior. On for . Then  



and so yielding the point On for



and so yielding the point On for



and so yielding the point On for



and so yielding the point Finally, we have (the minimum), and (the maximum).

(c) Find the hottest and coldest points on the metal plate whose temperature is given by

    Solution. Since is continuous and is closed and bounded, we know that the absolute maximum and minimum exist. We find that



and so the only critical point in the interior of is This is a saddle point because the discriminant of at is negative. The boundary of consists of four line segments and as follows. On and we have for which achieves a maximum at and a minimum at and Similarly, on and we have for which achieves its maximum at and and its minimum at We see that the hottest points are and and the coldest points are and

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Relative Extrema Of Functions
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
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