Extrema of two-variable function in bounded region

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Homework Help Overview

The discussion revolves around finding the absolute maxima and minima of the function T(x,y) = x² + xy + y² - 6x within a specified rectangular region defined by 0 ≤ x ≤ 5 and -3 ≤ y ≤ 3. Participants are exploring methods for identifying critical points and evaluating the function on boundaries.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessity of checking boundaries and whether to find extrema without considering the region first. There is mention of evaluating the function at critical points and comparing values to determine maxima and minima.

Discussion Status

Some participants have provided guidance on evaluating the function along the boundaries and suggested finding critical points within the interior of the region. There is an ongoing exploration of the relevance of the second derivative test and the importance of checking corner points.

Contextual Notes

Participants note the need to consider the corners of the rectangular region and the implications of directional derivatives for determining local extrema at these points. There is also a reference to the Karush-Kuhn-Tucker conditions in relation to corner evaluations.

xWaffle
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Homework Statement



Find absolute maxima and minima of the function in the given region:

T(x,y) = x2 + xy + y2 - 6x

Region: Rectangular plate given by: 0 ≤ x ≤ 5, -3 ≤ y ≤ 3

Homework Equations



First derivative test, fx =0, fy = 0
Second derivative test, fxxfyy - fxy2 = ?

The Attempt at a Solution



I know I need to check the boundaries of the region, which would be four lines connecting the endpoints (0, -3), (0, 3), (5, -3), (5, 3).

Do I also need to check for extrema the normal way (i.e., not caring about the region) and then see if any of those points lie within the region..? I have a general idea of what I'm supposed to be doing but I'm not 100% sure.
 
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On the boundaries you need to find the extrema of a function of a single variable. Inside the rectangle you need to find the extrema of the full expression. So just find the zeroes of the gradient and check which of those are inside the rectangle.
 
It isn't really necessary to use the second derivative test since whether you have a local max or min is not relevant. Evaluate the function at all possible critical points and compare to see which is largest and which smallest.

You have shown that the partial derivatives are never 0 in the interior of the region so need to check on boundaries. Since this is a square you have four lines as boundary.

1) x= 0. [itex]f(0, y)= y^2[/itex]. Find critical points of that for y from -3 to 3.
2) x= 5. [itex]f(5, y)= 25+ 5y+ y^2- 30= y^2+ 5y- 5[/itex].
3) y= -3. [itex]f(x, -3)= x^2- 3x+ 9- 6x= x^2- 9x+ 9[/itex].
4) y= 3. [itex]f(x, 3)= x^3+ 3x+ 9- 6x= x^2- 3x+ 9[/itex].

And, of course, you will need to check boundaries of those: the corners, (0, -3), (0, 3), (5, -3), and (5, 3). Find the values of the functions at all the critical points and see which is largest and which smallest.
 
xWaffle said:

Homework Statement



Find absolute maxima and minima of the function in the given region:

T(x,y) = x2 + xy + y2 - 6x

Region: Rectangular plate given by: 0 ≤ x ≤ 5, -3 ≤ y ≤ 3

Homework Equations



First derivative test, fx =0, fy = 0
Second derivative test, fxxfyy - fxy2 = ?

The Attempt at a Solution



I know I need to check the boundaries of the region, which would be four lines connecting the endpoints (0, -3), (0, 3), (5, -3), (5, 3).

Do I also need to check for extrema the normal way (i.e., not caring about the region) and then see if any of those points lie within the region..? I have a general idea of what I'm supposed to be doing but I'm not 100% sure.

You also need to worry about the corners (0,-3), etc. A corner is a local minimum if the directional derivative of f is ≥ 0 for all directions pointing into the region (or along the boundary) from that corner, and it is a local max if the directional derivative is ≤ 0. (These tests are, essentially, what the so-called Karush-Kuhn-Tucker conditions are all about.)

RGV
 

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