Find Absolute Extrema Over Region R

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    Absolute Extrema
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Discussion Overview

The discussion revolves around finding the absolute extrema of the function f(x, y) = 12 - 3x - 2y over a triangular region R defined by the vertices (2, 0), (0, 1), and (1, 2). Participants explore the steps involved in solving this problem, including the evaluation of the function at the boundaries and the vertices of the region.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Some participants suggest that graphing the vertices of the triangular region is a good initial step for solving the problem.
  • It is noted that the extrema of a differentiable function on a closed and bounded region must occur either in the interior, where partial derivatives are zero or do not exist, or on the boundary.
  • Participants calculate the function values at the endpoints of the boundary segments, indicating that the function is linear and its derivatives are never zero.
  • Specific calculations for the function along each boundary segment are provided, showing that extrema must occur at the endpoints of the segments.
  • Some participants emphasize that the maximum and minimum values can be found directly by evaluating the function at the vertices of the triangle.

Areas of Agreement / Disagreement

There is no explicit consensus on the necessity of graphing the points as a first step, as some participants agree it is a good start while others suggest it may not always be necessary. The discussion remains unresolved regarding the best approach to finding the extrema.

Contextual Notes

Participants mention that the partial derivatives of the function are constant and never equal to zero, which influences the search for extrema. The discussion includes detailed evaluations along the boundary segments, but does not resolve the necessity of graphing as a step.

harpazo
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Find the absolute extrema of the function over the region R. (In this case, R contains the boundaries.)

f (x, y) = 12 - 3x - 2y

R: The triangular region in the xy-plane with vertices (2, 0),
(0, 1), and (1, 2).

I need the steps to solve such a problem.
I was told to graph the 3 given points on the xy-plane as step 1. Is this true?
 
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Harpazo said:
Find the absolute extrema of the function over the region R. (In this case, R contains the boundaries.)

f (x, y) = 12 - 3x - 2y

R: The triangular region in the xy-plane with vertices (2, 0),
(0, 1), and (1, 2).

I need the steps to solve such a problem.
I was told to graph the 3 given points on the xy-plane as step 1. Is this true?

You could do that. You should realize that as the function is a plane, its extrema can only be somewhere on the boundary. So can you find the points on the boundary which maximise the function?
 
Harpazo said:
Find the absolute extrema of the function over the region R. (In this case, R contains the boundaries.)

f (x, y) = 12 - 3x - 2y

R: The triangular region in the xy-plane with vertices (2, 0),
(0, 1), and (1, 2).

I need the steps to solve such a problem.
I was told to graph the 3 given points on the xy-plane as step 1. Is this true?
That is typically a good start for any such problem but not all you need to do nor always necessary.

The absolute extrema of differentiable function on a closed and bounded region must occur in interior of the region, where the partial derivatives are 0 (or do not exist) or on the boundary. The partial derivatives here are [math]f_x= -3[/math] and [math]f_y= -2[/math]. The partial derivatives are never 0 so we must look on the boundary.

The boundary consists of three line segments:
1) The segment from (2, 0) to (0, 1). This has equation y= -x/2+ 1. f(x, -x/2+ 1)= 12- 3x- 2(-x/2+ 1)= 12- 3x+ x- 2= 10- 2x. The derivative of that is -2 which is never 0 so there is no extremum in the interior of this segment. Any extremum must be on the boundary- the end points, (2, 0) and (0, 1). f(2, 0)= 12- 6- 0= 6. f(0, 1)= 12- 0- 2= 10.

2) The segment from (2, 0) to (1, 2). This has equation y= -2x+ 4. f(x, -2x+ 4)= 12- 3x- 2(-2x+ 4)= 12- 3x+ 4x- 8= x+ 4. The derivative of that is 1 which is never 0 so there is no extremum in the interior of this segment. Any extremum must be on the boundary- the end points, (2, 0) and (1, 2). As before f(2, 0)= 6. f(1, 2)= 12- 3- 4= 5.

3) The segment from (0, 1) to (1, 2). This has equation y= x+ 1. f(x, x+ 1)= 12- 3x- 2(x+ 1)= 12- 3x- 2x- 2= 10- 5x. The derivative of that is -5 which is never 0 so there is no extremum in the interior of this segment. Any extremem must be on the boundary- the endpoints, (0, 1) and (1, 2). As before f(0, 1)= 10 and f(1, 2)= 5.

As Prove It pointed out, this function is linear so its derivatives are never 0. We could have just immediately evaluated the function at the vertices of the triangle. Those are f(0,1)= 10, f(2, 0)= 6 and f(1, 2)= 5. Of those, 10 is largest and 5 is the smallest so the maximum is 10, occurring at (0, 1) and 5 is the minimum, occurig at (1, 2).
 
HallsofIvy said:
That is typically a good start for any such problem but not all you need to do nor always necessary.

The absolute extrema of differentiable function on a closed and bounded region must occur in interior of the region, where the partial derivatives are 0 (or do not exist) or on the boundary. The partial derivatives here are [math]f_x= -3[/math] and [math]f_y= -2[/math]. The partial derivatives are never 0 so we must look on the boundary.

The boundary consists of three line segments:
1) The segment from (2, 0) to (0, 1). This has equation y= -x/2+ 1. f(x, -x/2+ 1)= 12- 3x- 2(-x/2+ 1)= 12- 3x+ x- 2= 10- 2x. The derivative of that is -2 which is never 0 so there is no extremum in the interior of this segment. Any extremum must be on the boundary- the end points, (2, 0) and (0, 1). f(2, 0)= 12- 6- 0= 6. f(0, 1)= 12- 0- 2= 10.

2) The segment from (2, 0) to (1, 2). This has equation y= -2x+ 4. f(x, -2x+ 4)= 12- 3x- 2(-2x+ 4)= 12- 3x+ 4x- 8= x+ 4. The derivative of that is 1 which is never 0 so there is no extremum in the interior of this segment. Any extremum must be on the boundary- the end points, (2, 0) and (1, 2). As before f(2, 0)= 6. f(1, 2)= 12- 3- 4= 5.

3) The segment from (0, 1) to (1, 2). This has equation y= x+ 1. f(x, x+ 1)= 12- 3x- 2(x+ 1)= 12- 3x- 2x- 2= 10- 5x. The derivative of that is -5 which is never 0 so there is no extremum in the interior of this segment. Any extremem must be on the boundary- the endpoints, (0, 1) and (1, 2). As before f(0, 1)= 10 and f(1, 2)= 5.

As Prove It pointed out, this function is linear so its derivatives are never 0. We could have just immediately evaluated the function at the vertices of the triangle. Those are f(0,1)= 10, f(2, 0)= 6 and f(1, 2)= 5. Of those, 10 is largest and 5 is the smallest so the maximum is 10, occurring at (0, 1) and 5 is the minimum, occurig at (1, 2).

Excellent reply!
 

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