Relative and Absolute Extrema Math 200 Week 6 - Friday Relative and Absolute Extrema

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

Relative Minima at x = -1 and x = 1 Math 200 Example: Second Derivative Test Consider the function f(x) = x4 - 2x2 f’(x) = 4x3 - 4x = 4x(x2 - 1) Critical points: x=-1,0,1 (because f’(-1)=0, f’(0)=0, f’(1)=0) f’’(x) = 12x2 - 4 f’’(-1) = 8 > 0 so f is concave up at x=-1 f’’(0) = -4 < 0 so f is concave down at x=0 f’’(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 points If 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

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

An easy Example (We already know the answer) Math 200 An easy Example (We already know the answer) Consider the paraboloid f(x,y) = x2 + y2 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)

Test the concavity of f at (0,0) by looking at D: 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 fxx(0,0) > 0 we have a relative minimum at (0,0)

Math 200

Example 1 Consider the function Math 200 Example 1 Consider the function Find the critical points by setting fx and fy 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

That’s one critical point: (0,0) Let’s do the same for y = 1: 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. If D(x0,y0) > 0 and fxx(x0,y0) < 0, then f has a relative maximum at (x0,y0) If D(x0,y0) > 0 and fxx(x0,y0) > 0, then f has a relative minimum at (x0,y0) If D(x0,y0) < 0, then f has a saddle point at (x0,y0) (x,y) fxx(x,y) fyy(x,y) fxy(x,y) D(x,y) Type (0,0) -2 -3 6 Relative Max (-1,1) 3 -6 -36 Saddle Point (1,1)

Math 200 Relative Maximum Saddle Points

Example 2 Find all relative extrema for f(x,y) = 4 + x3 + y3 - 3xy Math 200 Example 2 Find all relative extrema for f(x,y) = 4 + x3 + y3 - 3xy We have to be a little more clever here…solve the x-partial for y: Now plug that into the y-partial

Now we put together what we know: y = x2 and x = 0 or 1 y = (0)2 = 0 Math 200 Now we put together what we know: y = x2 and x = 0 or 1 y = (0)2 = 0 y = (1)2 = 1 Critical Points: (0,0), (1,1) Finally, we test the concavity of f at these points using D (x,y) fxx fyy fxy D Type (0,0) -3 -9 Saddle (1,1) 6 27 Rel. Min

Math 200 Saddle Point Relative Minimum

Absolute Extrema on a closed interval (From Calc 1) 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

A quick Calc 1 example: f(x) = x3 - 6x2 + 3 on [-1,5] Math 200 A quick Calc 1 example: f(x) = x3 - 6x2 + 3 on [-1,5] First we find the critical points f’(x) = 3x2 - 12x = 3x(x-4) Critical points: x = 0, 4 Evaluate f at x = -1, 0, 4, 5 f(-1) = -5 f(0) =3 f(4) = -29 f(5) = -22 Absolute Max: 3 at x = 0 Absolute min: -29 at x = 4

New stuff: Closed regions 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

Updated Extreme value Theorem 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

Example x2 = 4 - 4y2 Replacing x2 with 4 - 4y2 we get a function of y Math 200 Example Consider f(x,y) = x2 + y2 restricted to the elliptic region R: x2 + 4y2 ≤ 4 First we find the critical points for f inside R fx = 2x fy = 2y Critical 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 = 4 x2 = 4 - 4y2 Replacing x2 with 4 - 4y2 we get a function of y b(y) = 4 - 4y2 + y2 b(y) = 4 - 3y2

Find critical points: b’(y) = -6y y = 0 Evaluate and compare: b(0) = 4 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 - 3y2 Find critical points: b’(y) = -6y y = 0 Evaluate and compare: b(0) = 4 b(-1) = 1 b(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) = x2 + 4y2 - 2x2y + 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

f(x,y) = x2 + 4y2 - 2x2y + 4 f has three critical points: 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≤1 To restrict f to this boundary piece, we set y=-1 b(x) = f(x,-1) = x2 + 4(-1)2 - 2x2(-1) + 4 = 3x2 + 8 b’(x) = 6x Critical point: x = 0 b(0) = f(0,-1) = 8

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

f(x,y) = x2 + 4y2 - 2x2y + 4 Right: x = 1 and -1 ≤ y ≤ 1 Restrict f: Math 200 Right: x = 1 and -1 ≤ y ≤ 1 Restrict f: b(y) = f(1,y) = (1)2 + 4y2 - 2(1)2y + 4 = 4y2 - 2y + 5 b’(y) = 8y - 2 Critical point: y = 1/4 b(1/4) = f(1,1/4) = 19/4 Notice, 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) = 11

Critical points from boundary pieces: f(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)

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