Find Absolute Extrema Over Region R

[harpazo]

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?

 

[Prove It]

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?

 

[HallsofIvy]

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 \(\displaystyle f_x= -3\) and \(\displaystyle f_y= -2\). 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).

 

[harpazo]

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 \(\displaystyle f_x= -3\) and \(\displaystyle f_y= -2\). 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!