Math 233. Extrema of Functions of Two Variables Basics Theorem (Extreme Value Theorem) Let f be a continuous function of two variables x and y defined on a closed bounded region R in the xy-plane. Then f attains its maximum and minimum value on R. Example 1.(a) Let R = {(x, y) : 1/2 x 1, 0 y 1}. The function f(x, y) = 1/x 2 + y 3 is continuous on R so it has a maximum and minimum on R because R is a closed and bounded region. Let R = {(x, y) : 0 < x < 1, 0 y 1}. The function f(x, y) = 1/x 2 + y 3 is continuous on R, but R is not closed so there is no guaranted that f has a maximum and minimum (or even that f is bounded) on R. (Relative Extrema) Let f be a function defined on a region R containing (x 0, y 0 ). (a) The function f has a relative minimum at (x 0, y 0 ) if f(x, y) f(x 0, y 0 ) for all (x, y) in an open disk containing (x 0, y 0 ). (b) The function f has a relative maximum at (x 0, y 0 ) if f(x, y) f(x 0, y 0 ) for all (x, y) in an open disk containing (x 0, y 0 ). (Critical Points) Let f be defined on an open region R containing (x 0, y 0 ), then (x 0, y 0 ) is said to be a critical point of f if one of the following is true. (a) f x (x 0, y 0 ) = 0 and f y (x 0, y 0 ) = 0 (b) f x (x 0, y 0 ) or f y (x 0, y 0 ) does not exist. Example 2. The function f(x, y) = x 2 + y 2 and the function g(x, y) = x 2 + y 2 both have critical points at (0, 0), but for different reasons. Theorem. If f has a relative extremum at (x 0, y 0 ) on an open region R, then (x 0, y 0 ) is a critical point of f This follows from the one dimensional result by looking at the traces of f: If f(x, y 0 ) has a relative extremum at x 0, then either the derivative is 0, or does not exist. Similarly, if f(x 0, y) has a relative extremum at y 0, then either the derivative is 0 or does not exist. Theorem. (Second Derivative Test) Suppose f has continuous second partial derivatives on an open region containing a point (a, b) for which f x (a, b) = 0 and f y (a, b) = 0.
Consider the quantity d = f xx (a, b)f yy (a, b) [f xy (a, b)] 2 (a) If d > 0 and f xx (a, b) > 0, then f has a relative minimum at (a, b). (b) If d > 0 and f xx (a, b) < 0, then f has a relative maximum at (a, b). (c) If d < 0, then f has a saddle point at (a, b). (d) If d = 0, then the test is inconclusive. Some easy baseline examples to keep in mind with this theorem are the functions f(x, y) = x 2 +y 2, f(x, y) = x 2 y 2, f(x, y) = x 2 y 2 that have minimum, maximum and saddle at (0, 0) as confirmed by the test; however, the test is inconclusive on the functions f(x, y) = x 4 + y 4, f(x, y) = x 4 y 4 and f(x, y) = x 4 y 4 which respectively have minimum, maximum and saddle at (0, 0). Example 3. Find and classify the relative extrema of f(x, y) = 4xy x 4 y 4. Answer. Use the second derivative test to classify the extrema of f(x, y) = 4xy x 4 + y 4. First find the critcal points by solving f x = 4y 4x 3 = 0 and f y = 4x 4y 3 = 0. The first equation implies y = x 3, and we substitute this into the second equation to obtain 4x 4x 9 = 0, or 4x(1 x 8 ) = 0. This implies x = 0, x = 1, or x = 1. Therefore (0, 0) and ( 1, 1) and (1, 1) are the critical points. Now compute f xx = 12x 2, f yy = 12y 2, and f xy = 4 and recall, D = f xx f yy f 2 xy. At (0, 0), D = 0 16 = 0, and so f has a saddle point at (0, 0). At (1, 1), D = 144 16 > 0, and f xx < 0, and so f has a relative maximum at (1, 1). At ( 1, 1), D = 144 16 > 0 and f xx < 0, and so f has a relative maximum at ( 1, 1). The Extreme Value Theorem ensures a continuous function has a maximim and minimum on a closed bounded region. One process for finding the extrema is to first find critical points in the region, and then try to reduce the problem to a one variable problem on the boundary of the region. This is illustrated in the following: Example 4. Find the maximum and minimum values of the function f(x, y) = 3x 2 + 2y 2 4y over the region in the xy-plane bounded by the graphs of y = x 2 and y = 9. Answer. First, the critial point of f is where 6x = 0 and 4y 4 = 0, and so the critical point is (0, 1), and f(0, 1) = 2. Now find maxima and minima of f along the boundary of the region. When y = 9 and 3 x 3, f(x, 9) = 3x 2 + 2(81) 36 = 3x 2 + 126. This is a parabola opening upward, so its minimum is at the vertex x = 0, and the maxima are at the endpoints
x = ±3. Thus on this boundary, the minimum is f(0, 9) = 126, and the maxima are f(±3, 9) = 27 + 126 = 153. When y = x 2 and 3 x 3, then f(x, x 2 ) = 3x 2 + 2x 4 4x 2 = 2x 4 x 2. Setting the derivative equal to 0, we find 8x 3 2x = 0, or 2x(4x 2 1) = 0, so x = 0, or x = ± 1. Thus 2 along this curve, at the critical points f(0) = 0, f(± 1) = 1 1 = 1, and at the endpoints 2 8 4 8 f(±3) = 2(81) 9 = 153 (which should agree with endpoints of other boundary, as they are the same endpoints). Thus the maximum value that f takes on the region is f(±3, 9) = 153, and the minimum value f takes on the region is f(0, 1) = 2.
Practice Exercises on Extrema 1. Find and classify the relative extrema of f(x, y) = x 5 5xy + 5 2 y2. 2. Find and classify the critical points of f(x, y) = xy 8x 2 y 10xy 2. 3. Find the maximum and minimum values of the function f(x, y) = 3x 2 + 2y 2 8y + 5 over the region in the xy-plane bounded by the graphs of y = x 2 and y = 16.
Practice Exercises on Extrema with Solutions. 1. Find and classify the relative extrema of f(x, y) = x 5 5xy + 5 2 y2. Solution: First, we find the critical points f x = 5x 4 5y = 0 and f y = 5x+5y = 0. The second equation implies y = x; substituting this into the first implies 5x 4 5x = 0, and so x(x 3 1) = 0 which implies x = 0 or x = 1. Thus the critcial points are (0, 0) and (1, 1). Next, f xx = 20x 3, f yy = 5 and f xy = 5. Therefore, D = f xx f yy f 2 xy = 80x 3 25. At the critical point (0, 0), D = 25 < 0 and so f has a saddle at (0, 0). At the critical point (1, 1), D = 55 > 0, and f xx = 20 > 0, and so f has a relative minimum at (1, 1). 2. Find and classify the critical points of f(x, y) = xy 8x 2 y 10xy 2. Solution: To find the critical points we solve f x = 0 and f y = 0. Thus and f x = y 16xy 10y 2 = y(1 16x 10y) = 0 f y = x 8x 2 20xy = x(1 8x 20y) = 0 The first equation implies y = 0 or 16x + 10y = 1. We plug y = 0 into the second equation and obtain x(1 8x) = 0, and thus we obtain critical points (0, 0) and (1/8, 0). In the event 16x + 10y = 1, we substitute 10y = 1 16x into the second equation and so x(1 8x 2(1 16x)) = 0 = x(24x 1) = 0 and x = 0 implies y = 1/10, and x = 1/24 implies y = 1/30. Altogether the critical points are (0, 0), (1/8, 0), (0, 1/10) and (1/24, 1/30). The second partials needed for the second derivative test are f xx = 16y, f yy = 20x and f xy = 1 16x 20y. To apply the second derivative test for each critical point we compute D = f xx f yy f 2 xy At (0, 0): D(0, 0) = (0)(0) 1 2 < 0 and so f has a saddle at (0, 0).
At (1/8, 0): f xx = 0, f xy = 1 so D(1/8, 0) = 0 ( 1) 2 < 0, and so f has a saddle at (1/8, 0). At (0, 1/10): f yy = 0, f xy = 1 so D(0, 1/10) = 0 ( 1) 2 < 0, and so f has a saddle at (0, 1/10). At (1/24, 1/30): f xx = 16 30, f yy = 20 24 and f xy = 1 3 D(1/24, 1/30) = 4 9 1 9 > 0 and then Because D > 0 and f xx < 0, f has a local maximum at (1/24, 1/30). 3. Find the maximum and minimum values of the function f(x, y) = 3x 2 + 2y 2 8y + 5 over the region in the xy-plane bounded by the graphs of y = x 2 and y = 16. Solution: First, the critial point of f is where 6x = 0 and 4y 8 = 0, and so the critical point is (0, 2), and f(0, 2) = 3. (If you look carefully at the original function, you can conclude this is the absolute minimum of the function). Now find maxima and minima of f along the boundary of the region. When y = 16 and 4 x 4, f(x, 16) = 3x 2 + 2(16 2 ) (8)(16) + 5 = 389 + 3x 2. This is a parabola opening upward, so its minimum is at the vertex x = 0, and the maxima are at the endpoints x = ±4. Thus on this boundary, the minimum is f(0, 16) = 389, and the maxima are f(±4, 16) = 389 + 3(4 2 ) = 437. When y = x 2 and 4 x 4, then f(x, x 2 ) = 3x 2 + 2x 4 8x 2 + 5 = 5 5x 2 + 2x 4 Again, set the derivative equal to 0 and check the critical points (the endpoints of this curve are the same as the endpoints above). Thus setting the derivative equal to 0, we find 4(2)x 3 2(5)x = 0, or 2x(4x 2 5) = 0, so x = 0, or x = ± 5/4. Checking the values of f at these critical points and comparing to the values above we find the extreme values of f over the region are F (0, 2) = 3 (minimum) and f(±4, 16) = 437 (maximum)