Absolute Maxima and Minima Section 12.11

Transcription

1 Absolute Maxima and Minima Section 1.11 Definition (a) A set in the plane is called a closed set if it contains all of its boundary points. (b) A set in the plane is called a bounded set if it is contained in a disk (no matter how large the disk is). 1

2 Extreme Value Theorem If f(x,y) is continuous on a closed and bounded subset D of the plane, then on D the function f has an absolute maximum and an absolute minimum. Process of finding absolute minimum and maximum values on a closed bounded set 1. Find the interior critical points of f in D. Then calculate the values of f at these points.. Restrict the function on each piece of the boundary of D and then find the maximum of f on each piece. 3. Compare all the values found in the two parts above. Then the largest value among them is the absolute maximum of f, and the smallest value among them is the absolute minimum. Example (example 1.3 of the textbook) Temperature in degrees Celsius at each point (x,y) in the x + y 1 semicircular plate is given by y 1 T (x,y) = 16x xy + 0y

3 Find the hottest and coldest points of the plate. Solution 0 = T x = 3x y x 3y = 0 and 0 = T y = x + 80y and 3x + 10y = 0 (x,y) = (0,0) this point is on the boundary The value of T at this point is zero: T = 0. Next Step. We study the function on the boundary. The boundary consists of two parts: a semicircle which we call L 1, and a straight line segment 1 x 1 which we call L. Analysis on L 1 : For L 1 one might take y = 1 x, and then the restriction of the function T on the semicircle will be T = 16x x 1 x + 0(1 x ) = 0 x x 1 x But a better way of representing T on the semicircle is found by taking x = cost and y = sint : T = 16cos t costsint + 0sin t 0 t π = 8(1 + cost) 1sin(t) + 0(1 cos(t)) = 0 1cos(t) 1sin(t) This method is better because we don t have to deal with the square root. dt dt dt dt = (sint cost) = 0 sint cost = 0 sint = cost tant = 1 3

4 t = π and t = 5π t = π 8 and t = 5π 8 So here we have two other candidates: (x,y) = (cos π 8, sin π 8 ) T = 11 (x,y) = (cos 5π 8, sin 5π 8 ) T = 5 Analysis on L : On L we have y = 0 and 1 x 1. So on this part of the boundary, the temperature function T reduces to: T = 16x dt dx = 0 3x = 0 So, there is only one critical point, which corresponds to (0,0), at which we have T = 0. Finally, consider the corner points: Finally we must consider the the corner points, which ( 1,0) and (1,0). At both of these points we have T = 16. So, in summary, we have found these important points: candidate (x, y) T(x, y) (0,0) 0 absolute min (coldest) (cos π 8, sin π 8 ) 11 (cos 5π 8, sin 5π 8 ) 5 absolute max (hottest) ( 1, 0) 16 (1, 0) 16

5 Constraint Opyimization from section 1.11 Example (example 1.35 of the textbook) Find the points in the first octant part of the plane 6x + 3y + z = 6 closest to the point (,6,7). Assume that the lines of intersection of the plane with the coordinate planes are part of the surface. Solution Let (x,y,z) represent an arbitrary point of the plane in the first octant. Then the distance to be minimized is (x ) + (y 6) + (z 7). We want to minimize this function subject to the constraints 6x + 3y + z = 6 x 0 y 0 z 0 Since the square root function is an increasing function, this minimization problem has the same (x,y,z) solution as the minimization problem involving the minimization of (x ) + (y 6) + (z 7). However, this second problem is easier to work with as there is no square root involved. Now the problem becomes: min (x ) + (y 6) + (z 7) 6x + 3y + z = 6 subject to x 0 y 0 z 0 In the constraint optimization problems we get rid of one of the unknowns; for example we find z = 6 6x 3y and substitute to reduce the problem to the following problem: min (x ) + (y 6) + ( 6 6x 3y 7) 6x + 3y 6 subject to x 0 y 0 Equivalently: 5

6 min f(x,y) = (x ) + (y 6) (6x + 3y + ) 6x + 3y 6 subject to x 0 y 0 We label the three parts of boundary by L 1, L, and L 3, as shown. Step 1. Search for the critical points: f x = 0 f y = 0 (x ) + 3 (6x + 3y + ) = 0 (y 6) (6x + 3y + ) = 0 13 x + 9 y = 17 9 x y = 15 Solving this system, one gets (x,y) = ( 10 acceptable. 61, ), but this point is not in the region therefore not Analysis on L 1 : On L 1 the problem reduces to: f = 16 + (y 6) (3y + ) 0 y Differentiate and set equal to zero: df dy = 0 (y 6) (3y + ) = 0 y = 6 5 f = 80 The point that gives this value is the point (0, 6 5, 3 5 ). Analysis on L : On L the problem reduces to: f = (x ) (3x + 11) 0 x 1 6

7 Differentiate and set equal to zero: df dx = 0 (x ) + 3 (3x + 11) = 0 x = not acceptable Analysis on L 3 : On L 3 we have x + y =, therefore y = x ; so by substituting we reduce the problem to: f = (x ) + ( x 6) (6x + 6 6x + ) = (x ) + (x + ) x 1 Differentiate and set equal to zero: df dx = 0 (x ) + 8(x + ) = 0 x = 5 not acceptable Finally, consider the corner points: Finally we must consider the the corner points, which (0,0), (1,0) and (0,). At these points we have the following values for the function f respectively: f (0,0) = 39 f (1,0) = 9 f (0,) = 81 These three points are (0, 0, 6 ), (1, 0, 0), (0,, 0) So, in summary, we have found these important points: candidate (x,y,z) square of the distance (0, 6 5, 3 5 ) 80 absolute min (0, 0, 6 ) 39 (1, 0, 0) 9 (0,, 0) 81 So, the minimum distance will be 80 and it happens at the point (0, 6 5, 3 5 ) 7