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Extrema of Functions of Two Variables

Extrema of Functions of Two Variables. Absolute Extrema and Relative Extrema. Absolute Extrema and Relative Extrema. Consider the continuous function f of two variables, defined on a closed bounded region R . The values f ( a , b ) and f ( c , d ) such that

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Extrema of Functions of Two Variables

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  1. Extrema of Functions of Two Variables

  2. Absolute Extrema and Relative Extrema

  3. Absolute Extrema and Relative Extrema Consider the continuous function f of two variables, defined on a closed bounded region R. The values f(a, b) and f(c, d) such that f(a, b) ≤ f(x, y) ≤ f(c, d) (a, b) and (c, d) are in R. for all (x, y) in R are called the minimum and maximum of f in the region R, as shown in Figure 13.64. R contains point(s) at which f(x, y)is a minimum and point(s) at which f(x, y) is a maximum.

  4. Absolute Extrema and Relative Extrema A region in the plane is closed if it contains all of its boundary points. The Extreme Value Theorem deals with a region in the plane that is both closed and bounded. A region in the plane is called bounded if it is a subregion of a closed disk in the plane.

  5. Absolute Extrema and Relative Extrema

  6. Absolute Extrema and Relative Extrema A minimum is also called an absolute minimum and a maximum is also called an absolute maximum. As in single-variable calculus, there is a distinction made between absolute extrema and relative extrema.

  7. Absolute Extrema and Relative Extrema To say that f has a relative maximum at (x0, y0) means that the point (x0, y0, z0) is at least as high as all nearby points on the graph of z = f(x, y). Similarly, f has a relative minimum at (x0, y0) if (x0, y0, z0) is at least as low as all nearby points on the graph. Relative extrema

  8. Absolute Extrema and Relative Extrema To locate relative extrema of f, you can investigate the points at which the gradient of f is 0 or the points at which one of the partial derivatives does not exist. Such points are called critical points of f.

  9. Absolute Extrema and Relative Extrema If f is differentiable and then every directional derivative at (x0, y0) must be 0. This implies that the function has a horizontal tangent plane at the point (x0, y0), as shown. Relative minimum Relative maximum

  10. Absolute Extrema and Relative Extrema It appears that such a point is a likely location of a relative extremum. This is confirmed by Theorem 13.16.

  11. Example 1 Find the critical points of f(x, y) = xy(x – 2)(y + 3). f(x, y) is differentiable at all points of R2, so the critical points occur only at points where fx(x, y) = fy(x, y) = 0. f(x, y) = xy(xy + 3x – 2y – 6) = x2y2 + 3x2y – 2xy2 – 6xy fx(x, y) = 2xy2 + 6xy – 2y2 – 6y = 2xy(y + 3) – 2y(y + 3) = (2xy – 2y)(y + 3) = 2y(x – 1)(y + 3)

  12. fy(x, y) = 2x2y + 3x2 – 4xy – 6x = x2(2y + 3) – 2x(2y + 3) = (x2 – 2x)(2y + 3) = x(x – 2)(2y + 3) We must now identify all (x, y) pairs that satisfy both equations. The first equation is satisfied if and only if y = 0, x = 1, or y = -3. We consider each of these cases……

  13. Substituting y = 0, the second equation is 3x(x – 2) = 0, which has solutions x = 0 and x = 2. So, (0, 0) and (2, 0) are critical points. • Substituting x = 1, the second equation is –(2y + 3) = 0, which has the solution y = -3/2. So, (1, -3/2) is a critical point. • Substituting y = -3, the second equation is -3x(x – 2) = 0, which has roots x = 0 and x = 2. So, (0, -3) and (2, -3) are critical points. We find that there are five critical points: (0, 0), (2, 0), (1, -3/2), (0, -3), and (2, -3). Some of these critical points may correspond to local maximum or minimum values. We return to this example and a complete analysis.

  14. The Second Partials Test

  15. The Second Partials Test To find relative extrema you need only examine values of f(x, y) at critical points. However, as is true for a function of one variable, the critical points of a function of two variables do not always yield relative maxima or minima. Some critical points yield saddle points, which are neither relative maxima nor relative minima.

  16. The Second Partials Test As an example of a critical point that does not yield a relative extremum, consider the surface given by f(x, y) = y2 – x2 Hyperbolic paraboloid as shown. Saddle point at (0, 0, 0): fx(0, 0) = fy(0, 0) = 0

  17. The Second Partials Test At the point (0, 0), both partial derivatives are 0. The function f does not, however, have a relative extremum at this point because in any open disk centered at (0, 0) the function takes on both negative values (along the x-axis) and positive values (along the y-axis). So, the point (0, 0, 0) is a saddle point of the surface.

  18. The Second Partials Test

  19. Example 2 Use the Second Partials Test to classify the critical points of f(x, y) = x2 + 2y2 – 4x + 4y + 6. The four second partials are…… fx = 2x – 4, fy = 4y + 4, fxx = 2, fxy = fyx = 0, fyy = 4 Setting both fx and fy equal to zero yields the single critical point (2, -1). D(2, -1) = fxxfyy – (fxy)2 = 8 > 0 and fxx(2, -1) = 2 > 0 By the Second Partials Test, f has a local minimum at (2, -1); the value of the function at that point is f(2, -1) = 0.

  20. Return to Example 1 Use the Second Partials Test to classify the critical points of f(x, y) = xy(x – 2)(y + 3). We determined the critical points of f are (0, 0), (2, 0), (1, -3/2), (0, -3), and (2, -3). The derivatives needed to use the Second Partials Tests are…… fx(x, y) = 2y(x – 1)(y + 3), fy(x, y) = x(x – 2)(2y + 3), fxx(x, y) = 2y(y + 3), fxy(x, y) = 2(2y + 3)(x – 1), fyy(x, y) = 2x(x – 3)

  21. The surface described by f has one local maximum at (1, -3/2), surrounded by four saddle points.

  22. The structure of the surface may also be visualized by plotting the level curves of f.

  23. Example 4 Apply the Second Partials Test to f(x, y) = 2x4 + y4. The critical points of f satisfy the conditions…… fx(x, y) = 8x3 = 0 and fy(x, y) = 4y3 = 0, so, the sole critical point is (0, 0). The second partials evaluated at (0, 0) are…… fxx(0, 0) = fxy(0, 0) = fyy(0, 0) = 0. We see that D(0, 0) = 0, and the Second Partials Test is inconclusive. While the bowl-shaped surface……

  24. described by f has a local minimum at (0, 0), the surface also has a broad flat bottom, which makes the local minimum “invisible” to the Second Partials Test.

  25. Example 5 Apply the Second Partials Test to f(x, y) = 2 – xy2. The critical points of this function satisfy…… fx(x, y) = -y2 = 0 and fy(x, y) = -2xy = 0. The solutions of these equations have the form (a, 0), where is a real number. It is easy to check that the second partial derivatives evaluated at (a, 0) are…… fxx(a, 0) = fxy(a, 0) = 0 and fyy(a, 0) = -2a. Therefore, D(a, 0) = 0, and the Second Partials Test is inconclusive.

  26. This shows that f has a flat ridge above the x-axis that the Second Partials Test is unable to classify.

  27. Absolute Maximum and Minimum Values

  28. Example 6 Find the absolute maximum and minimum values of f(x, y) = x2 + y2 – 2x + 2y + 5 on the set R = {(x, y): x2 + y2 ≤ 4} (the closed disk centered at (0, 0) with radius 2). We begin by locating the critical points and the local maxima and minima. The critical points satisfy the equations…… fx(x, y) = 2x – 2 = 0 and fy(x, y) = 2y + 2 = 0, which have the solution x = 1 and y = -1. The value of the function at this point is f(1, -1) = 3.

  29. We now determine the maximum and minimum values of f on the boundary of R, which is a circle of radius 2 described by the parametric equations…… x = 2 cos θ, y = 2 sin θ, for 0 ≤ θ ≤ 2π. Substituting x and y in terms of θ into the function f, we obtain a new function g(θ) that gives the values of f on the boundary of R: g(θ) = (2 cos θ)2 + (2 sin θ)2 – 2(2 cos θ) + 2(2 sin θ) + 5 = 4(cos2 θ + sin2 θ) – 4 cos θ + 4 sin θ + 5 = -4 cos θ + 4 sin θ + 9.

  30. Finding the maximum and minimum boundary values is now a one-variable problem. The critical points of g satisfy…… g’(θ) = 4 sin θ + 4 cos θ = 0, or tan θ = -1. Therefore, g has critical points θ = -π/4 and θ = (3π)/4, which correspond to the points (√2, -√2) and (-√2, √2), respectively. The function values at these points are……

  31. Having completed the first two steps of this procedure, we have three function values to consider: • f(1, -1) = 3 (critical point), • f(√2, -√2) = 9 – 4√2 ≈ 3.3 (boundary point), and • f(-√2, √2) = 9 + 4√2 ≈ 14.7 (boundary point) The greatest value, f(-√2, √2) = 9 + 4√2, is the absolute maximum value, and it occurs at a boundary point. The least value, f(1, -1) = 3, is the absolute minimum value, and it occurs at an interior point.

  32. Also revealing is the plot of the level curves of the surface with the boundary of R superimposed.

  33. As the boundary of R is traversed, the values of f vary, reaching a maximum value at θ = (3π)/4, or (-√2, √2), and a minimum value at θ = -π/4, or (√2, -√2).

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