Local Extrema Previously we have taken the partial derivative of a function f(x, y). But those partial derivatives were themselves functions and so we can take their partial derivatives. Local Maximums and Local Minimums of Functions f(x, y) has a local maximum at the point (x0, y0) if f(x0, y0) ¥ f(x, y) for all points (x, y) near (x0, y0) f(x, y) has a local minimum at the point (x0, y0) if f(x0, y0) § f(x, y) for all points (x, y) near (x0, y0) Recall that the gradient vector points in a direction where the function increases. Suppose that the point (x0, y0) is a local maximum of the function f(x, y) which is not on the boundary of the domain. If the vector “f(x0, y0) is defined and non-zero, then we can increase f(x, y) by moving in the direction of “f(x0, y0). But we just said that the point (x0, y0) is a local maximum, thus there is no direction that we can travel in to make the function larger. Hence, we must have that “f(x0, y0) = 0. Similarly, if (x0, y0) were a local minimum, it would follow that “f(x0, y0) = 0 as well. Using our results from above, we also have a useful relation between local maxima and local minima to first-order partial derivatives. Local Extrema Theorem If f(x, y) has a local maximum or a local minimum at (x0, y0) and the first-order partial derivatives of f(x, y) exist there, then fx(x0, y0) = 0 and fy(x0, y0) = 0. This leads us to the definition of a critical point for a function. As is the case with one-variable calculus, we allow for the possibility that a point is neither a local maximum nor a local minimum. Critical Points of a Function A point (x0, y0) is said to be a critical point of the function f(x, y) if the gradient, “f(x0, y0), is either 0 or undefined. 1Example 1: Find and analyze the critical point(s) of f(x, y) = x2 + xy + y2. Solution: fx(x, y) = 2x + y and fy(x, y) = x + 2y. Setting these equations equal to 0, we have two equations: 2x + y = 0 and x + 2y = 0 We can solve for x and y by multiplying the second equation by -2 and adding the equations. Doing this, we have that -3y = 0, and so y = 0. Plugging that into the first equation, we see that x = 0. Thus, the only critical point is (0, 0). Notice that f(0, 0) = 0. Thus, if f(x, y) is always positive or zero near (0, 0), then we would have that (0, 0) is a local minimum. If f(x, y) is always negative or zero near (0, 0), then we would have a local maximum. Looking at a sketch of the graph (Figure 1 below), it would appear that (0, 0, 0) is a local minimum. Figure 1: Sketch of f(x, y) = x2 + xy + y2 We can confirm our suspicion by completing the square on the function. That is, we have that 22213(, )242fxy x xy y x y y . Notice that this is a sum of two squares, so it is always greater than or equal to zero. Thus, the critical point is a local (and in fact, global) minimum. In Example 1, solving for x and y was fairly straightforward. We had two linear equations with two unknowns. We were able to solve for x and y by solving the equations simultaneously or by the method substitution. 2In some instances, though, a bit more algebra is required. Often times, it is best to substitute the result of one of equations into the other equation to solve for the variables, as the next example shows. Example 2: Find the local extrema of the function 2323(, ) 2fxy x xy y  . Solution: fx(x, y) = 2x – 2y and fy(x, y) = –2x + 2y2. Setting these equal to 0, we have the following equations: 2x – 2y = 0 and –2x + 2y2 = 0 Solving for 2x in both equations, we have 2x = 2y and 2x = 2y2. Setting the two equations equal to each other, we have that 2y = 2y2. Solving for y, we have that y = 0 or y = 1. If y = 0, then x = 0. And if y = 1, then x = 1. So, we end up with the critical points (0, 0) and (1, 1). We can use a contour diagram to help classify critical points. Let us begin by looking at some critical points and classify them. 0.250.51234566778899910101010-1 1x-11yA Figure 2: A local minimum at the point A = (0, 0) In Figure 2, the level curves are circular around the point A. This suggests that we are dealing with a local maximum or a local minimum. Moreover, as we move away from the point A, the level curves are getting larger. This means that A is a local minimum. 3-6-6-4-4-2-2-2-1-1-0.5-0.5-0.25-0.25-1 1x2yBC Figure 3: Local maximums at the points B = (-1, 0) and C = (1, 2) In Figure 3, the level curves are circular around the points B and C. This suggests that we are dealing with local maxima or a local minima. Moreover, as we move away from the points, the level curves are getting smaller (more negative). This means that B and C are local maxima. -3.5-3.5-3-3-2.5-2.5-2-2-1.5-1.5-1-1-0.5-0.50000.50.5111.51.5222.52.5333.53.5xyD Figure 4: A saddle point at the point D = (0, 0) In Figure 4, the level curves are perpendicular at the point D. As we move in the positive x-direction, the values of the level curves get larger. As we move in the positive y-direction, the values of the level curves get smaller. This means that D is a saddle point. From our figures above, we have the following classification of critical points based upon their level curves. 4Determining Critical Points from a Contour Diagram Suppose that (x0, y0) is a critical point of the function f(x, y). - If the contours around (x0, y0) are circular, (x0, y0) is either a local maximum or a local minimum. + If the values of the level curves decrease as you move away from the point (x0, y0), then (x0, y0) is a local maximum. + If the values of the level curves increase as you move away from the point (x0, y0), then (x0, y0) is a local minimum. - If two contours meet at a perpendicular angle, (x0, y0) is a saddle point. Example 3: In Figure 5 (below), classify the points P, Q, and R as local maximum, local minimum, saddle points, or neither. 0123.13.13.73.7444.256-1 1x-11yPQR Figure 5: Contour Diagram of f(x, y) = 4 + x3 + y3 – 3xy Solution: Since the level curves are perpendicular at P, we see that P is a saddle point. Around the point Q, we notice that the level curves are circular. Also, as we move away from the point, the values of the level curves are increasing. Thus, Q is a local minimum. None of the …