Multiple variables - check understanding

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SUMMARY

The discussion focuses on finding the minimum and maximum values of the function g(x,y) = 3x² - x - y + y² over the domain D = [0,1] x [0,1]. Participants outline a systematic approach that includes evaluating the function at the boundary points (0,0), (0,1), (1,0), and (1,1), as well as the critical point (1/6, 1/2) where the gradient is zero. A total of nine points are identified for evaluation, including critical points from the boundaries. The discussion also raises a question about evaluating points when the gradient yields no critical points.

PREREQUISITES
  • Understanding of multivariable calculus, specifically optimization techniques.
  • Familiarity with gradient and critical points in functions of two variables.
  • Knowledge of evaluating functions over a closed interval.
  • Ability to differentiate functions with respect to multiple variables.
NEXT STEPS
  • Study the method of Lagrange multipliers for constrained optimization.
  • Learn about evaluating functions at boundary conditions in multivariable calculus.
  • Explore the implications of critical points and their significance in optimization.
  • Investigate the behavior of gradients in functions with no critical points.
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Students and professionals in mathematics, particularly those studying calculus and optimization techniques, as well as educators looking for examples of multivariable function analysis.

Niles
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Homework Statement



A function g(x,y) = 3x^2-x-y+y^2. I have to find the minimum and maximum of the function on D = [0,1] x [0,1].

The Attempt at a Solution



First I have to check the ends; I mean (0,0), (0,1), (1,0) and (1,1). I also have to check the points, where the gradient is zero: (1/6,1/2).

I insert x = 0, x = 1, y = 0 and y = 1 in g(x,y) and differentiate and equal to zero - then I have 4 new poins to check.

I also insert x = 1/6 and y = 1/2 and do the same as above.

So totally I have 11 points to check?
 
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Hmm, I get nine points:

1. The critical point for the open domain (0,1)*(0,1), that is (1/6,1/2)

2. The four critical points for the function evaluated at the four boundary lines x=0, y=0, x=1 and y=1

3. The four end points of the boundary lines, i.e, the corners (0,0), (0,1), (1,0), (1,1).
 
Ok, so I do not need to insert the critical points and evaluate them (diff. and equal zero)?

What if the gradient is something like (2x-1,1) - then only x = 0 and there's no critical points. Do I have to evaluate x = 0?

Thank you for a quick respond!
 

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