Second Partials Test

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.

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

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Second Partials Test
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
http://www.libraryofmath.com/second-partials-test.html