Results 1 to 3 of 3

Math Help - Quick question on extremas

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    2

    Question Quick question on extremas

    In lectures we did questions like (in the simplest form) f(x,y,z) = x + 2y + z subject to a constraint, eg. x^2 + y^2 + z^2 = 48 Find the critical points, value and determine the max/min

    In the exam papers some of the questions have the constraint with <= instead of just =. Does this make a big difference to how I approach the question?

    Also, when I get the critical values, how do I determine which points are a max/min? Not specifically the example above but in general.

    Lastly does the term "on a closed disk" in place of constraint hold any significance?

    Thanks in advance for any help...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by GenericNameA View Post
    In lectures we did questions like (in the simplest form) f(x,y,z) = x + 2y + z subject to a constraint, eg. x^2 + y^2 + z^2 = 48 Find the critical points, value and determine the max/min

    In the exam papers some of the questions have the constraint with <= instead of just =. Does this make a big difference to how I approach the question?

    Also, when I get the critical values, how do I determine which points are a max/min? Not specifically the example above but in general.

    Lastly does the term "on a closed disk" in place of constraint hold any significance?

    Thanks in advance for any help...
    The question you posted now is to be solved with Lagrange multipliers.

    When you have \leq instead of =, for example, e^{x+y} on x^2+y^2 \leq 1 (this is on a closed disk) then you need to approach this problem by setting the partial to zero and checking the boundary values for extreme values.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,313
    Thanks
    1291
    Quote Originally Posted by GenericNameA View Post
    In lectures we did questions like (in the simplest form) f(x,y,z) = x + 2y + z subject to a constraint, eg. x^2 + y^2 + z^2 = 48 Find the critical points, value and determine the max/min

    In the exam papers some of the questions have the constraint with <= instead of just =. Does this make a big difference to how I approach the question?
    I doubt that you did not cover this in class- just in a different section that you are thinking of. If you have, for example, the problem that says "minimize f(x,y,z) on x^2+ y^2+ z^2= 48 you would probably, as ThePerfectHacker suggested, use Lagrange multipliers.

    But if the problem is "minimize f(x,y,z) on x^2+ y^2+ z^2\le 48, that's a completely different problem. You would first minimize f(x,y,z) on the interior of the sphere by finding [itex]\nabla f[/itex], finding where it is 0, and seeing if any such points are inside the sphere. Then you would look for additional max and min on the surface of the sphere- that is by using the constraint x^2+ y^2+ Z^2= 48.
    Actually, for the example you give, f(x,y,z)= x+ 2y+ z, \nabla f= \vec{i}+ 2\vec{j}+ \vec{k} is never 0 so the problem really reduces to the constraint on the surface of the sphere.

    Also, when I get the critical values, how do I determine which points are a max/min? Not specifically the example above but in general.
    If you are talking about global min or max on a closed bounded set, just calculate the value of the function at each critical value and see which is largest and which smallest. For determining whether critical values are local max or min or neither, surely your text has the "second derivative test": "If [tex]D= f_xxf_yy- (f_xy)^2> 0 and either f_xx or f_yy is positive, the critical point is a minimum. If D> 0 and either f_xx or f_yy is negative, the critical point is a maximum. If D< 0 the critical point is a saddle point. If D= 0 this test doesn't work."

    Lastly does the term "on a closed disk" in place of constraint hold any significance?

    Thanks in advance for any help...
    That is the same as the ball above. Look for critical points inside the disk and then look for additional critical points on the boundary of the disk.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Questions about relative extremas?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 25th 2008, 08:51 PM
  2. Critical Numbers & Extremas
    Posted in the Calculus Forum
    Replies: 10
    Last Post: November 9th 2008, 03:36 PM
  3. Quick Lin Alg question
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: September 4th 2008, 08:49 PM
  4. Global Extremas
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 4th 2007, 02:11 PM
  5. Quick's quick question
    Posted in the Number Theory Forum
    Replies: 22
    Last Post: July 9th 2006, 04:38 PM

Search Tags


/mathhelpforum @mathhelpforum