MHB Critical Points & Extrema of Multivariable Function

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SUMMARY

The discussion focuses on finding critical points and testing for relative extrema of the multivariable function f(x, y) = x^(2/3) + y^(2/3). The critical point identified is (0, 0), where the Second Partials Test fails, resulting in a determinant d of 0. The function has a cusp at (0, 0), which is classified as an absolute minimum, while it lacks any relative minima or maxima, as it is unbounded elsewhere.

PREREQUISITES
  • Understanding of multivariable calculus concepts, particularly critical points.
  • Familiarity with the Second Partials Test for classifying extrema.
  • Knowledge of partial derivatives and their computation.
  • Concept of cusps in the context of function behavior.
NEXT STEPS
  • Study the application of the Second Partials Test in greater detail.
  • Learn about the classification of critical points in multivariable functions.
  • Explore the concept of cusps and their implications in calculus.
  • Investigate other methods for finding extrema in multivariable functions, such as Lagrange multipliers.
USEFUL FOR

Students and professionals in mathematics, particularly those studying multivariable calculus, as well as educators seeking to clarify concepts related to critical points and extrema.

harpazo
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Find the critical points and test for relative extrema. List the critical points for which the Second Partials Test fails.

f (x, y) = x^(2/3) + y^(2/3)

Solution:

f_x = 2/[3 (x)^1/3]

f_y = 2/[3 (y)^1/3]

f_xx = -2/[9 x^(4/3)]

f_yy = -2/[9 y^(4/3)]

f_xy = 0

I set f_x and f_y to 0 and found the critical point to be
(0, 0).

To find (0, 0, 0), I evaluated f (x, y) at the point (0, 0).

Can you please tell me what (0, 0, 0) represents here? I am confused about the critical point (0, 0) and the point in space (0, 0, 0). Are they the same point?

Is this ok so far?

d = -2/[9 x^(4/3)]*-2/[9 x^(4/3)] - [0]^2

I then evaluated d at the point (0, 0) and the result is 0.
This means the test fails.

The textbook goes on to say that there is absolute minimum in this case.

Is any of this correct? Why do we have absolute minimum here and not relative minimum?
 
Physics news on Phys.org
The function:

$$f(x,y)=x^{\frac{2}{3}}+y^{\frac{2}{3}}$$

has no critical or stationary points. It has a "cusp" at $(x,y)=(0,0)$, which serves as the absolute or global minimum, and is otherwise unbounded (no global maximum).
 
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