Find the absolute maximum and minimum vales of the set D.
F(x,y) = x^2 + y^2 - 2x, D is the closed triangular region with vertices (2,0), (0,2), and (0,-2).
I know that I have to find Fx(x,y) and Fy(x,y) but I'm not sure where to go from there. Any help is appreciated.
Fx(x,y) = x -1; Fy(x,y) = 2y
To find the absolute maximum and minimum values of F(x, y) = x^2 + y^2 - 2x on the closed triangular region D with vertices (2, 0), (0, 2), and (0, -2), you can follow these steps:
1. Find the critical points: Set both partial derivatives equal to zero and solve for x and y:
Fx(x, y) = x - 1 = 0 --> x = 1
Fy(x, y) = 2y = 0 --> y = 0
So, the critical point is (1, 0).
2. Check the boundary points: To find the absolute maximum and minimum on the boundary of D, evaluate F(x, y) at all three vertices (2, 0), (0, 2), and (0, -2). Also, evaluate F(x, y) along each side of the triangle and compare the results.
For each vertex, substitute the x and y values into F(x, y):
F(2, 0) = 2^2 + 0^2 - 2(2) = 4
F(0, 2) = 0^2 + 2^2 - 2(0) = 4
F(0, -2) = 0^2 + (-2)^2 - 2(0) = 4
For each side, parameterize the line segment and substitute the parameter values into F(x, y). Since D is a closed triangular region, we can consider the three sides:
- The line segment from (2, 0) to (0, 2):
The equation of the line is x + y = 2.
Substitute x = 2 - t and y = t into F(x, y):
F(2-t, t) = (2-t)^2 + t^2 - 2(2-t) = 2t^2 - 2t.
We need to find the maximum and minimum of F(2-t, t) as t ranges from 0 to 2.
Take the derivative of F(2-t, t) with respect to t and set it equal to zero:
F'(2-t, t) = 4t - 2 = 0 --> t = 1/2
Substitute t = 1/2 into F(2-t, t):
F(2-(1/2), 1/2) = (3/2)^2 + (1/2)^2 - 2(2 - (1/2)) = 1/2
Therefore, the absolute maximum is F(2-(1/2), 1/2) = 1/2 and the absolute minimum is F(2-(1/2), 1/2) = 1/2.
- The line segment from (0, 2) to (0, -2):
The equation of the line is x = 0.
Substitute x = 0 and y = t into F(x, y):
F(0, t) = 0^2 + t^2 - 2(0) = t^2
We need to find the maximum and minimum of F(0, t) as t ranges from -2 to 2.
Substitute t = -2 and t = 2 into F(0, t):
F(0, -2) = 0^2 + (-2)^2 - 2(0) = 4
F(0, 2) = 0^2 + 2^2 - 2(0) = 4
Therefore, the absolute maximum is F(0, 2) = 4 and the absolute minimum is F(0, -2) = 4.
3. Compare the values obtained from the critical point and the boundary points to find the absolute maximum and minimum values:
The absolute maximum value is 4 (obtained at (0, 2) and (0, -2)), and the absolute minimum value is 1/2 (obtained at (1, 0) and (2-1/2, 1/2)).
Therefore, the absolute maximum value of F on D is 4, and the absolute minimum value is 1/2.