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

2. 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.

3. 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

4. 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

5. 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

6. 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…

7. 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)

8. 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)

9. Math 200

10. 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

11. 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)

12. 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)

13. Math 200
Relative Maximum
Saddle Points

14. 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

15. 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

16. Math 200
Saddle Point
Relative Minimum

17. 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

18. 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

19. 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

20. 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

21. 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 (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

22. 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)

23. Math 200

24. 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

25. 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

26. 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

27. 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

28. 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: (1,-1) and (-1,-1)
Absolute min: (0,0)

29. 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)