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Maxima/Minima in a conceptual equation

  1. Sep 19, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the stationary points of the function:
    z = ax^2 + by^2 + c
    For each of the following sub-cases, identify any maxima and minima.
    i) a > 0, b > 0
    ii) a < 0, b < 0
    iii) a and b of opposite signs.


    2. Relevant equations
    z = ax^2 + by^2 + c


    3. The attempt at a solution
    Z'(x) = 2ax = 0
    Z'(y) = 2by = 0

    Z'(x) = Z'(y) => 2ax = 2by
    Solving for y: Y = ax/b
    Solving for x: X = by/a

    Z''(xx) = 2a
    Z''(yy) = 2b
    Z''(xy) = 0

    Checking the condition [Z''(xx) * Z''(yy)] - (Z''(xy))^2 --> (2a*2b) - (0)^2 = 4ab.

    For i) where a and b are > 0 --> 4ab > 0 and 2a > 0 ---> Minimum
    For ii) where a and b are < 0 --> 4ab > 0 and 2a < 0 ---> Maximum
    For iii) where a and b are opposite --> 4ab < 0 --> Saddle Point

    Have I done this correctly? Because I can help thinking that I have forgotten something. I would really appriciate any help and input with regards to my attempted solution! - Thanks!
     
  2. jcsd
  3. Sep 19, 2011 #2

    dynamicsolo

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    Homework Helper

    This looks to be correct (though your partial derivative notation should be something more like zxxzyy - ( zxy )2 , for instance).

    This surface is an easy one to check up on. If you alternately set either x = 0 or y = 0 , you find that the cross-sections on the yz- and xz-planes are parabolas. When the signs of a and b match, both parabolic cross-sections open in the same direction. When the signs are opposite, the parabolas are opening in opposite directions.

    So having both a and b positive gives us an (elliptic) paraboloid which opens upward (in the positive-z direction), meaning that the critical point at (x, y) = (0, 0) is a minimum. Both a and b being negative has the paraboloid opening downward, so (0, 0) becomes a maximum. For a and b having opposite signs, the surface is a hyperbolic paraboloid and (0, 0) is the saddle point.
     
  4. Sep 19, 2011 #3
    Hey dynami, and thanks for your help!

    I'm afraid I am not as confident as you are when it comes to shapes in the xy plane so I dont think I have a chance of "seeing" what you just described without actually plotting the graphs manually on a piece of paper hehe.

    However, what you just described, does that always hold? I.e. if a and b are both positive, the paraboloid will always open upwards? What I mean is, can I apply that reasoning to any equation (with actual numbers replacing a and b) and "see" the form straight away before even doing the check with the double partial derivatives?

    Again, thanks for your help!
     
  5. Sep 19, 2011 #4

    dynamicsolo

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    If we set y = 0 for the surface, we get z = ax2 + c ; this is the cross-section in the xz-plane and is the equation of a parabola. Since x2 is always positive or zero, if a is positive, z increases toward "positive infinity" as | x | becomes larger and larger, so the parabola opens upward (the value of c just shifts the parabola along the z-axis, so it has no effect on the "opening direction"). Similarly, if a is negative, z goes toward "negative infinity" as | x | becomes very large; we have a "downward-opening" parabola. We can apply this in the same way for the yz-plane (setting x = 0 ).

    So we are applying analytic geometry, which works pretty well for us here because we had a good idea of how parabolas work. For more complicated functions of two variables, it can become a lot harder to "picture" the function, so analysis by partial derivatives is more effective. And plainly, when we go to functions of more than two dimensions, this kind of analysis is largely all we can go by, since our ability to "see" what the function does becomes very limited...
     
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