Analyzing Critical Points of f(x,y)=Ax2+E

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SUMMARY

The critical points of the function f(x,y) = Ax² + E, where A and E are constants, are located along the y-axis at (0,n) for any real number n. The first partial derivative with respect to x yields 2Ax = 0, indicating that x must be 0 for critical points. The second partial derivatives reveal that Df = 0 at these points, which complicates the determination of local extrema. However, graphical analysis using Mathematica confirms that if A > 0, the critical points are global minima, while if A < 0, they are global maxima, indicating that these critical points are saddle points.

PREREQUISITES
  • Understanding of partial derivatives and critical points in multivariable calculus.
  • Familiarity with the concepts of local maxima, local minima, and saddle points.
  • Knowledge of the Hessian determinant and its role in classifying critical points.
  • Experience with graphing functions using tools like Mathematica.
NEXT STEPS
  • Study the classification of critical points using the Hessian matrix in multivariable calculus.
  • Learn how to use Mathematica for visualizing functions and their critical points.
  • Explore the implications of different values of A on the shape of the function f(x,y) = Ax² + E.
  • Investigate the geometric interpretation of functions with parabolic cross-sections.
USEFUL FOR

Students and educators in calculus, mathematicians analyzing multivariable functions, and anyone interested in understanding the behavior of parabolic surfaces in mathematical contexts.

xokaitt
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Homework Statement



Let f(x,y)=Ax2+E where A and E are constants. What are the critical points of f(x,y)? Determine whether the critical points are local maxima, local minema, or saddle points.


2. The attempt at a solution

First I found the first partial derivatives with respect to x and y
[tex]\partial[/tex]f/[tex]\partial[/tex]x=2Ax
[tex]\partial[/tex]f/[tex]\partial[/tex]y=0
[tex]\Rightarrow[/tex] 2Ax=0,
[tex]\Rightarrow[/tex] x=0 for any constant A.

Therefore, all points lying on the y-axis are critical points.
(i.e. C.P.'s = (0,n), n[tex]\in[/tex]R.)

Now, we have to find the second partial's with respect to x and y.
[tex]\partial[/tex]2f/[tex]\partial[/tex]x2=2A
[tex]\partial[/tex]2f/[tex]\partial[/tex]y2=0
and
[tex]\partial[/tex]2f/[tex]\partial[/tex]x[tex]\partial[/tex]y=0

Therefore Df=([tex]\partial[/tex]2f/[tex]\partial[/tex]x2)([tex]\partial[/tex]2f/[tex]\partial[/tex]y2)-([tex]\partial[/tex]2f/[tex]\partial[/tex]x[tex]\partial[/tex]y)2 at (0,n) , n[tex]\in[/tex]R.
[tex]\Rightarrow[/tex] Df=(2A)(0)-(0)2=0

This is where I get stuck. Now that Df=0, how do I determine whether or not the critical pts are local extrema or saddle pts?

From plotting the function on Mathematica, I know that these critical points are in fact saddle points, but I don't know how to mathematically state that.

Thanks!
 
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It seems to me that you are making this much harder than it needs to be by not sketching a graph of this surface. Since y doesn't appear explicitly in the formula for the function, this surface is a cylinder with parabolic cross section, and with its axis of symmetry in the direction of the y-axis. IOW, the surface looks something like a trough. If A > 0, the trough opens upward, and all critical points are global minima. If A < 0, the trough opens downward, and all critical points are global maxima. All critical points lie on a line that is parallel to the y-axis.
 

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