RELATIVE AND ABSOLUTE EXTREMA MATH 200 GOALS Be able to use - - PowerPoint PPT 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) = 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

• f 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

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…

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:

• fx(x, y) = 2x

fy(x, y) = 2y = ⇒

• 2x = 0

2y = 0 = ⇒

• x = 0

y = 0 ▸ So, we have one critical point: (0,0)

MATH 200

▸ Test the concavity of f at (0,0) by looking at D: D(x, y) = fxx(x, y)fyy(x, y) − (fxy(x, y))2 ▸ To do so we need the second order partial derivatives: fxx(x, y) = 2; fyy(x, y) = 2; fxy(x.y) = 0 ▸ 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

MATH 200

EXAMPLE 1

▸ Consider the function f(x, y) = x2y3 − 3 2y2 − x2 ▸ Find the critical points by setting fx and fy equal to zero

• fx(x, y) = 2xy3 − 2x

fy(x, y) = 3x2y2 − 3y

• 2xy3 − 2x = 0

3x2y2 − 3y = 0 ▸ In the first equation, we can factor out a 2x 2x(y3 − 1) = 0 = ⇒ x = 0, y = 1

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

3x2y2 − 3y = 0 = ⇒ 3(0)2y2 − 3y = 0 −3y = 0 y = 0 ▸ That’s one critical point: (0,0) ▸ Let’s do the same for y = 1: 3x2y2 − 3y = 0 = ⇒ 3x2(1)2 − 3(1) = 0 3x2 − 3 = 0 x2 − 1 = 0 x = ±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.

fxx = 2y3 − 2; fyy = 6x2y − 3; fxy = 6xy2

(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) 3 6

• 36

Saddle Point

D(x, y) = fxx(x, y)fyy(x, y) − (fxy(x, y))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)

MATH 200

SADDLE POINTS RELATIVE MAXIMUM

MATH 200

EXAMPLE 2

▸ Find all relative extrema for f(x,y) = 4 + x3 + y3 - 3xy

• fx(x, y) = 3x2 − 3y

fy(x, y) = 3y2 − 3x = ⇒

• 3x2 − 3y = 0

3y2 − 3x = 0 ▸ We have to be a little more clever here…solve the x-partial

for y:

3y = 3x2 = ⇒ y = x2 ▸ Now plug that into the y-partial 3(x2)2 − 3x = 0 3x4 − 3x = 0 3x(x3 − 1) = 0 x = 0, 1

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

fxx(x, y) = 6x fyy(x, y) = 6y fxy(x, y) = −3

(x,y) fxx fyy fxy D Type (0,0) 0 0 -3 -9 Saddle (1,1) 6 6 -3 27 Rel. Min

D = fxxfyy − (fxy)2

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) = 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 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 3x4 + 4x2y + y4 ≤ 4 3x4 + 4x2y + y4 < 4

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) = 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

MATH 200

▸ Now we can apply the EVT to b(y)

• n 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

• ccurs at (0,0)

▸ The absolute maximum is 4 which

• ccurs 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

• fx(x, y) = 2x − 4xy

fy(x, y) = 8y − 2x2

• 2x − 4xy = 0

8y − 2x2 = 0 ▸ Factor 2x in the x-partial: 2x(1 − 2y) = 0 = ⇒ x = 0, y = 1/2 ▸ Now we plug x=0 and y=1/2 into the y-partial x = 0 : 8y = 0 y = 0 y = 1/2 : 4 − 2x2 = 0 x = ± √ 2

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 (0, 0), ( √ 2, 1/2), (− √ 2, 1/2)

f(x,y) = x2 + 4y2 - 2x2y + 4

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

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 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) 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)