How is the Minimum Value of a Multivariable Function Determined in a Quadrant?

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SUMMARY

The discussion focuses on determining the minimum value of the multivariable function f(x,y) = x^3 - 3x^2 - 6xy + 7y + y^2 within the constraints x ≥ 0 and y ≥ 0. It is established that f attains its minimum either at a stationary point or along the boundaries of the quadrant. Critical points identified include (7, 35/2) and (1, -1/2), with the latter being invalid due to constraints. The Hessian determinant indicates that (7, 35/2) is a saddle point, confirming the need to analyze boundary behavior for minimum determination.

PREREQUISITES
  • Understanding of multivariable calculus, specifically critical points and Hessian matrices.
  • Familiarity with the method of Lagrange multipliers for constrained optimization.
  • Knowledge of boundary behavior analysis in optimization problems.
  • Proficiency in differentiating functions of multiple variables.
NEXT STEPS
  • Study the method of Lagrange multipliers for constrained optimization problems.
  • Learn about the classification of critical points using the Hessian matrix.
  • Research boundary behavior analysis in multivariable functions.
  • Explore the concept of boundedness in optimization within specific domains.
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Students and professionals in mathematics, particularly those focusing on calculus, optimization, and mathematical analysis. This discussion is beneficial for anyone tackling multivariable function optimization within defined constraints.

MeMoses
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Homework Statement


f(x,y) = x^3 - 3x^2 - 6xy + 7y + y^2, x>=0, y>=0
i) Explain why f attains its minimum value on the quadrant.
ii) Find the critical points and classify them


Homework Equations


df/dx = 3x^2 - 6x -6y
df/dy = -6x + 7 +2y

d^2f/dx^2 = 6x-6
d^2f/dy^2 = 2
d^2f/(dxdy) = -6

The Attempt at a Solution


It's been awhile since I've done problems like this. Hopefully I am making some sense.
I'm not sure about i). Couldn't x=-infinity and y=-1 yield -infinity?
For ii) I get the critical points to be (7, 35/2) and (1, -1/2), but (1, -1/2) is not in the constraints. fxx is positive and the determinant of the Hessian at the first point is negative so it is a saddle point. If the constraints weren't there, how would I figure out the other point since fxx=0?
iii) I'm not sure the best way to go about this. Help would be appreciated.
Thanks
 
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MeMoses said:

Homework Statement


f(x,y) = x^3 - 3x^2 - 6xy + 7y + y^2, x>=0, y>=0
i) Explain why f attains its minimum value on the quadrant.
ii) Find the critical points and classify them


Homework Equations


df/dx = 3x^2 - 6x -6y
df/dy = -6x + 7 +2y

d^2f/dx^2 = 6x-6
d^2f/dy^2 = 2
d^2f/(dxdy) = -6

The Attempt at a Solution


It's been awhile since I've done problems like this. Hopefully I am making some sense.
I'm not sure about i). Couldn't x=-infinity and y=-1 yield -infinity?
For ii) I get the critical points to be (7, 35/2) and (1, -1/2), but (1, -1/2) is not in the constraints. fxx is positive and the determinant of the Hessian at the first point is negative so it is a saddle point. If the constraints weren't there, how would I figure out the other point since fxx=0?
iii) I'm not sure the best way to go about this. Help would be appreciated.
Thanks

If f(x,y) does attain a minimum in {x,y ≥ 0} it does so either at an interior point (i.e., a stationary point) or on the boundary ({x=0} or {y=0}). You can check f along the two boundary lines {x=0,y≥0} and {y=0,x≥0}. Then, the only remaining question is whether f is bounded from below in the first quadrant. If I were doing the question I would check whether f is bounded from below on the non-negative x and y axes, and if it is bounded from below for points of the form (x,k*x) with k > 0 and x ≥ 0.

RGV
 

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