Review for Exam 2. Section 14 - Michigan State University
Review for Exam 2.
Sections 13.1, 13.3. 14.1-14.7. 50 minutes. 5 problems, similar to homework problems. No calculators, no notes, no books, no phones. No green book needed.
Section 14.7
Example
(a) Find all the critical points of f (x, y ) = 12xy - 2x3 - 3y 2. (b) For each critical point of f , determine whether f has a local
maximum, local minimum, or saddle point at that point.
Solution: (a) f (x, y ) = 12y - 6x2, 12x - 6y = 0, 0 , then,
x2 = 2y , y = 2x, x(x - 4) = 0. There are two solutions, x = 0 y = 0, and x = 4 y = 8. That is, there are two critical points, (0, 0) and (4, 8).
Section 14.7
Example
(a) Find all the critical points of f (x, y ) = 12xy - 2x3 - 3y 2. (b) For each critical point of f , determine whether f has a local
maximum, local minimum, or saddle point at that point.
Solution: (b) Recalling f (x, y ) = 12y - 6x2, 12x - 6y , we compute
fxx = -12x, fyy = -6, fxy = 12.
D(x, y ) = fxx fyy - (fxy )2 = 144
x -1
2
,
Since D(0, 0) = -144 < 0, the point (0, 0) is a saddle point of f .
Since D(4, 8) = 144(2 - 1) > 0, and fxx (4, 8) = (-12)4 < 0, the point (4, 8) is a local maximum of f .
Section 14.7
Example
Find the absolute maximum and absolute minimum of f (x, y ) = 2 + xy - 2x - 1 y 2 in the closed triangular region with
4 vertices given by (0, 0), (1, 0), and (0, 2).
Solution: We start finding the critical points inside the triangular region.
1 f (x, y ) = y - 2, x - y = 0, 0 , y = 2, y = 2x.
2 The solution is (1, 2). This point is outside in the triangular region given by the problem, so there is no critical point inside the region.
Section 14.7
Example
Find the absolute maximum and absolute minimum of f (x, y ) = 2 + xy - 2x - 1 y 2 in the closed triangular region with
4 vertices given by (0, 0), (1, 0), and (0, 2). Solution: We now find the candidates for absolute maximum and minimum on the borders of the triangular region. We first record the boundary vertices:
(0, 0) f (0, 0) = 2, (1, 0) f (1, 0) = 0, (0, 2) f (0, 2) = 1.
Section 14.7
Example
Find the absolute maximum and absolute minimum of f (x, y ) = 2 + xy - 2x - 1 y 2 in the closed triangular region with
4 vertices given by (0, 0), (1, 0), and (0, 2).
Solution: The horizontal side of the triangle, y = 0, x (0, 1). Since
g (x) = f (x, 0) = 2 - 2x, g (x) = -2 = 0.
there are no candidates in this part of the boundary. The vertical side of the triangle is x = 0, y (0, 2). Then,
g (y ) = f (0, y ) = 2 - 1 y 2,
1 g (y ) = - y = 0,
4
2
so y = 0 and we recover the point (0, 0).
Section 14.7
Example
Find the absolute maximum and absolute minimum of f (x, y ) = 2 + xy - 2x - 1 y 2 in the closed triangular region with
4 vertices given by (0, 0), (1, 0), and (0, 2). Solution:
The hypotenuse of the triangle y = 2 - 2x, x (0, 1). Then,
g (x)
=
f
(x ,
2
-
2x )
=
2
+
x (2
-
2x )
-
2x
-
1 (2
-
2x )2,
4
= 2 + 2x - 2x2 - 2x - (x2 - 2x + 1),
= 1 + 2x - 3x2.
Then,
g
(x )
=
2 - 6x
=
0
implies
x
=
1 3
,
hence
y
=
4 3
.
The
candidate is
1 3
,
4 3
.
Section 14.7
Example
Find the absolute maximum and absolute minimum of f (x, y ) = 2 + xy - 2x - 1 y 2 in the closed triangular region with
4 vertices given by (0, 0), (1, 0), and (0, 2).
Solution:
Recall that we have obtained a candidate point
1 3
,
4 3
.
We
evaluate f at this point,
14
4 2 1 16 4
f , =2+ - - = .
33
9 3 49 3
Recalling that f (0, 0) = 2, f (1, 0) = 0, and f (0, 2) = 1, the absolute maximum is at (0, 0), and the minimum is at (1, 0).
Section 14.6
Example
(a) Find the linear approximation L(x, y ) of the function f (x, y ) = sin(2x + 3y ) + 1 at the point (-3, 2).
(b) Use the approximation above to estimate the value of f (-2.8, 2.3).
Solution: (a) L(x, y ) = fx (-3, 2) (x + 3) + fy (-3, 2) (y - 2) + f (-3, 2). Since fx (x, y ) = 2 cos(2x + 3y ) and fy (x, y ) = 3 cos(2x + 3y ),
fx (-3, 2) = 2 cos(-6 + 6) = 2, fy (-3, 2) = 3 cos(-6 + 6) = 3, f (-3, 2) = sin(-6 + 6) + 1 = 1.
the linear approximation is L(x, y ) = 2(x + 3) + 3(y - 2) + 1.
Section 14.6
Example
(a) Find the linear approximation L(x, y ) of the function f (x, y ) = sin(2x + 3y ) + 1 at the point (-3, 2).
(b) Use the approximation above to estimate the value of f (-2.8, 2.3).
Solution: (b) Recall: L(x, y ) = 2(x + 3) + 3(y - 2) + 1. Now, the linear approximation of f (-2.8, 2.3) is L(-2.8, 2.3), and
L(-2.8, 2.3) = 2(-2.8 + 3) + 3(2.3 - 2) + 2 = 2(0.2) + 3(0.3) + 1 = 2.3.
We conclude L(-2.8, 2.3) = 2.3.
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- solution 7 university of california berkeley
- solving des by separation of variables
- 5 introduction to harmonic functions
- unit 5 implicit di erentiation related rates implicit
- 5 2 limits and continuity
- covariance and correlation math 217 probability and
- review for exam 2 section 14 michigan state university
- limits and continuity partial derivatives
- chapter 4 jointly distributed random variables
- first examples ucsd mathematics home
Related searches
- michigan state university job postings
- michigan state university philosophy dept
- michigan state university online degrees
- michigan state university employee discounts
- michigan state university employee lookup
- michigan state university employee portal
- michigan state university employee salaries
- michigan state university admissions
- michigan state university employee benefits
- michigan state university website
- michigan state university employee directory
- michigan state university deadline