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

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  • #1
MeMoses
129
0

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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  • #2
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
 

1. What is a multivariable extrema?

A multivariable extrema is a point on a multivariable function where the function reaches its maximum or minimum value. It can occur at a specific point or along a certain path on the function.

2. How is a multivariable extrema different from a single variable extrema?

A single variable extrema occurs on a function with only one independent variable, while a multivariable extrema occurs on a function with multiple independent variables. This means that the function can have multiple maximum or minimum points.

3. What are the methods for finding multivariable extrema?

The most common methods for finding multivariable extrema are using partial derivatives, the gradient vector, and the Hessian matrix. These methods involve finding critical points, where the partial derivatives are equal to zero, and then using the second derivative test to determine if they are maximum or minimum points.

4. Can multivariable extrema occur at a point where the partial derivatives do not exist?

Yes, multivariable extrema can occur at a point where the partial derivatives do not exist. These points are known as singular points and require further analysis to determine if they are maximum, minimum, or saddle points.

5. How are multivariable extrema used in real-world applications?

Multivariable extrema are used in many fields of science and engineering, such as physics, economics, and computer science. They can be used to optimize processes, minimize costs, and maximize efficiency. For example, in physics, multivariable extrema can be used to find the path of least resistance or the trajectory with the highest velocity. In economics, they can be used to determine the optimal production levels for a company.

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