Let f(x,y)= x^2+3y-3xy and R is the triangular region with vertices (0, 0), (2, 0) and (2, 2). Find the interior and boundary points only at which the absolute extrema of f(x, y) can occur
To find the absolute extrema of the function f(x, y) = x^2 + 3y - 3xy over the triangular region R, we need to evaluate the function at the interior points and along the boundary of the region.
First, let's find the interior points. These are points within the triangular region where we can find the critical points of the function. To find the critical points, we need to take the partial derivatives of f(x, y) with respect to x and y and set them equal to zero.
∂f/∂x = 2x - 3y = 0
∂f/∂y = 3 - 3x = 0
Solving these equations simultaneously, we get x = 1 and y = 2/3. So the point (1, 2/3) is a critical point within the triangular region R.
Now let's find the boundary points where the absolute extrema may occur. The boundary of the triangular region R is made up of three line segments:
1. From (0, 0) to (2, 0)
2. From (2, 0) to (2, 2)
3. From (2, 2) to (0, 0)
We need to evaluate the function f(x, y) at the endpoints of these line segments to find the absolute extrema.
1. For the line segment from (0, 0) to (2, 0):
Substituting y = 0 in the function f(x, y), we get f(x, 0) = x^2.
Evaluating at the endpoints, f(0, 0) = 0^2 = 0, and f(2, 0) = 2^2 = 4.
2. For the line segment from (2, 0) to (2, 2):
Substituting x = 2 in the function f(x, y), we get f(2, y) = 4 + 3y - 6y = 4 - 3y.
Evaluating at the endpoints, f(2, 0) = 4 - 3(0) = 4, and f(2, 2) = 4 - 3(2) = -2.
3. For the line segment from (2, 2) to (0, 0):
We can parametrize this line segment as (2-t, 2t) where 0 <= t <= 1.
Substituting these values in the function f(x, y), we get f(2-t, 2t) = (2-t)^2 + 3(2t) - 3(2-t)(2t).
Simplifying, we get f(2-t, 2t) = t^2 - 2t + 4.
Evaluating at the endpoints, f(2, 2) = 0^2 - 2(0) + 4 = 4, and f(0, 0) = 1^2 - 2(1) + 4 = 3.
Now we have evaluated the function at all the interior and boundary points, and we can compare these values to find the absolute extrema.
The absolute extrema are as follows:
- The maximum value is 4 and occurs at points (2, 0) and (2, 2) on the boundary.
- The minimum value is -2 and occurs at the point (2, 2) on the boundary, and the point (1, 2/3) in the interior.
Therefore, the interior and boundary points at which the absolute extrema of f(x, y) can occur are (2, 0), (2, 2), and (1, 2/3).