Relative and Absolute Extrema - PowerPoint Presentation

Relative and Absolute Extrema Math 200 Week 6 - Friday

Math 200 Goals Be able to use partial derivatives to find critical points (possible locations of maxima or minima). Know how to use the Second Partials Test for functions of two variables to determine whether a critical point is a relative maximum , relative minimum , or a saddle point . Be able to solve word problems involving maxima and minima. Know how to compute absolute maxima and minima on closed regions.

Math 200 From Calc 1 Given a function f(x), how do we find its relative extrema ? Find the critical points: f’(x) = 0 f’ is undefined Do either the first or second derivative test First derivative test: make a sign chart for f’ Second derivative test: look at the concavity of f at each critical point

Math 200 Example: Second Derivative Test Consider the function f(x) = x 4 - 2x 2 f’(x) = 4x 3 - 4x = 4x(x 2 - 1)Critical points: x=-1,0,1 (because f’(-1)=0, f’(0)=0, f’(1)=0)f’’(x) = 12x2 - 4f’’(-1) = 8 > 0 so f is concave up at x=-1f’’(0) = -4 < 0 so f is concave down at x=0f’’(1) = 8 > 0 so f is concave up at x=1 Relative Maximum at x = 0 Relative Minima at x = -1 and x = 1

Math 200 NEW STUFF The process for finding relative (local) extrema for functions of three variables follows the second derivative test from calc 1 pretty closely…but has a few more moving parts Step 1: Find critical points Places where we might have a relative extremum Step 2: Test the concavity of the function at the critical pointsIf f is concave up at a critical point, it’s a relative min If f is concave down at a critical point, it’s a relative max

Math 200 Critical Points: We say that (x 0 ,y 0 ) is a critical point for f provided that f x(x0,y0 ) = 0 and fy (x 0,y0 ) = 0Define D = fxxfyy - (fxy)2If D(x0 ,y0) > 0 and fxx(x0 ,y0) < 0, then f has a relative maximum at (x0,y0)If D(x 0 ,y 0 ) > 0 and f xx (x 0 ,y 0 ) > 0 , then f has a relative minimum at (x 0 ,y 0 ) If D(x 0 ,y 0 ) < 0 , then f has a saddle point at (x 0 ,y 0 ) If D(x 0 ,y 0 ) = 0, then the test fails…

Math 200 An easy Example (We already know the answer) Consider the paraboloid f(x,y) = x 2 + y 2 We already know (hopefully) that f has a relative minimum at (0,0), but let’s show this using the second derivative test. Find the critical points: So, we have one critical point: (0,0)

Math 200 Test the concavity of f at (0,0) by looking at D: To do so we need the second order partial derivatives: Since these are all constant, the calculation is pretty easy: D(0,0) = (2)(2) - 0 = 4 > 0 Since D(0,0) > 0 and f xx (0,0) > 0 we have a relative minimum at (0,0)

Math 200

Math 200 Example 1 Consider the function Find the critical points by setting f x and f y equal to zero In the first equation, we can factor out a 2x This does not mean (0,1) is a critical point! We can tell because it doesn’t work in the y-derivative

Math 200 x = 0 makes the x-partial zero for any y-value. Let’s plug that into the y-partial and see what happens That’s one critical point: (0,0) Let’s do the same for y = 1: That’s two more critical points: (-1,1) and (1,1)

Math 200 So far, we’ve found that f has three critical points. Now we want to test the concavity of f using D. (x,y) f xx (x,y) f yy (x,y) f xy(x,y) D(x,y) Type (0,0)-2 -3 06 Relative Max(-1,1) 0 3 -6 -36

Math 200 Saddle Points Relative Maximum

Math 200 Example 2 Find all relative extrema for f(x,y) = 4 + x 3 + y 3 - 3xy We have to be a little more clever here…solve the x-partial for y: Now plug that into the y-partial

Math 200 Now we put together what we know: y = x 2 and x = 0 or 1 y = (0) 2 = 0 y = (1) 2 = 1Critical Points: (0,0), (1,1)Finally, we test the concavity of f at these points using D (x,y) fxx fyyfxy DType (0,0) 0 0-3 -9 Saddle (1,1) 6 6 -3 27

Math 200 Relative Minimum Saddle Point

Math 200 Absolute Extrema on a closed interval (From Calc 1) Extreme value theorem : If a function f(x) is continuous on a closed interval [a,b], then f is guaranteed to have both a maximum value and a minimum value on [a,b] How we use the EVT for a continuous f on [a,b]: Find critical points on [a,b] Evaluate f(x) at each critical point inside [a,b] as well as at the endpoints (i.e. f(a) and f(b)) Compare f-values and identify maximum and minimum

Math 200 A quick Calc 1 example: f(x) = x 3 - 6x 2 + 3 on [-1,5] First we find the critical points f’(x) = 3x 2 - 12x = 3x(x-4)Critical points: x = 0, 4Evaluate f at x = -1, 0, 4, 5f(-1) = -5f(0) =3f(4) = -29f(5) = -22 Absolute min: -29 at x = 4 Absolute Max: 3 at x = 0

Math 200 New stuff: Closed regions For functions of two variables, we need to talk about closed and bounded regions rather than closed intervals Compare closed regions and open intervals

Math 200 Updated Extreme value Theorem If f(x,y) is continuous on a closed and bounded region R, then it must attain both a maximum value and a minimum value on that region. How to apply the EVT: Find all critical points for f inside the region R Restrict f to the boundary of R Apply the single-variable EVT to f restricted to the boundary Compare the values of f at the critical points inside R with the absolute extrema on the boundary

Math 200 Example Consider f(x,y) = x 2 + y 2 restricted to the elliptic region R: x 2 + 4y 2 ≤ 4First we find the critical points for f inside Rfx = 2xfy = 2yCritical point: (0,0)It’s certainly in R since 02 + 4(0)2 = 0 ≤ 4 We want to restrict f to the boundary of R Boundary: x2 + 4y2 = 4x 2 = 4 - 4y2Replacing x2 with 4 - 4y2 we get a function of y b(y) = 4 - 4y 2 + y2b(y) = 4 - 3y2

Math 200 Now we can apply the EVT to b(y) on the interval [-1,1] Why [-1,1]? Because the ellipse extends from y=-1 to y=1. b(y) = 4 - 3y 2 Find critical points: b’(y) = -6y y = 0 Evaluate and compare: b(0) = 4b(-1) = 1b(1) = 1 Compare these values with what we get for f at the critical point (0,0) f(0,0) = 0 So the absolute minimum is 0, which occurs at (0,0)The absolute maximum is 4 which occurs on the boundary of R when y=0 What is x when y=0 on the boundary?x2 = 4 - 4y2 x = -2 or 2 So, the absolute maximum occurs at (-2,0) and (2,0)

Math 200

Math 200 Another Example Find the absolute extrema for f(x,y) = x 2 + 4y 2 - 2x 2 y + 4 on the square S = {(x,y):-1≤x≤1 and -1≤y≤1} First, we’ll find the critical points in R Factor 2x in the x-partial: Now we plug x=0 and y=1/2 into the y-partial

Math 200 f has three critical points: But only (0,0) is in S! Evaluate: f(0,0) = 4 Now we need to see what happens on the boundary . S is a square, so we have to look at each side separately Bottom of square: y=-1 and -1≤x≤1To restrict f to this boundary piece, we set y=-1b(x) = f(x,-1) = x2 + 4(-1)2 - 2x2(-1) + 4 = 3x2 + 8b’(x) = 6x Critical point: x = 0 b(0) = f(0,-1) = 8 f(x,y) = x 2 + 4y2 - 2x2y + 4

Math 200 Top: y = 1 and -1 ≤ x ≤ 1 Restrict f: b(x) = f(x,1) = x 2 + 4(1) 2 - 2x 2 (1) + 4 = 8 - x 2b’(x) = -2xCritical point: x = 0b(0) = f(0,1) = 8Left: x = -1 and -1 ≤ y ≤ 1Restrict f: b(y) = f(-1,y) = (-1)2 + 4y2 - 2(-1)2y + 4 = 4y2 - 2y + 5b’(y) = 8y - 2Critical point: y = 1/4b(1/4) = f(-1,1/4) = 19/4 f(x,y) = x2 + 4y 2 - 2x2y + 4

Math 200 Right: x = 1 and -1 ≤ y ≤ 1 Restrict f: b(y) = f(1,y) = (1) 2 + 4y 2 - 2(1) 2 y + 4 = 4y2 - 2y + 5b’(y) = 8y - 2Critical point: y = 1/4b(1/4) = f(1,1/4) = 19/4Notice, we didn’t test the endpoints on the boundary pieces, so we should do so now. The endpoints are the four corners of the square: (1,1), (1,-1), (-1,1), (-1,-1)f(1,1) = 7, f(1,-1) = 11, f(-1,1) = 7, f(-1,-1) = 11f(x,y) = x2 + 4y2 - 2x2y + 4

Math 200 We have a lot to compare Critical points in S: f(0,0) = 4 Critical points from boundary pieces: f(0,-1) = 8, f(0,1) = 8, f(-1,1/4) = 19/4, f(1,1/4) = 19/4 Endpoints of boundary pieces: f(1,1) = 7, f(1,-1) = 11, f(-1,1) = 7, f(-1,-1) = 11 Absolute max: 11 @ (1,-1) and (-1,-1) Absolute min: 4 @ (0,0)f(x,y) = x2 + 4y2 - 2x2y + 4

Math 200 Absolute minimum of 4 at (0,0) Absolute Maximum of 11 at (1,-1) and (-1,-1)