bmb11e_ppt_15_3 lecture notes for students.ppt

Barnett/Ziegler/Byleen Business Calculus 11e 1
Learning Objectives for
Section 15.3 Maxima and Minima
The student will be able to
identify critical points and
maxima and minima of
functions.

Local Maxima and Minima
We want to examine local maxima and minima for functions
of the form z = f (x, y). We are going to extend the second-
derivative test previously developed for y = f (x).
Definition: f (a, b) is a local maximum if there exists a
circular region about (a, b) such that f (a, b)  f (x, y) for all
(x, y) in the region.
f (a, b) is a local minimum if there exists a circular region
about (a, b) such that f (a, b)  f (x, y) for all (x, y) in the
region.

First Test for Local Extremum
Theorem:
Let f (a, b) be a local extremum for the function f. If both fx
and fy exist at (a, b), then
fx (a, b) = 0 and f y (a, b) = 0.
Example: z = x2
+ y2
. This is a paraboloid with vertex at the
origin and opening upward. Consider the vertex (0, 0).
f x(x, y) = 2x and f x(0, 0) = 0,
f y(x, y) = 2y and f y(0, 0) = 0.

First Test
(continued)
Unfortunately the converse of Theorem 1 is not true. That is, if
fx(a, b) = 0 and fy(a, b) = 0,
it does not necessarily follow that there is a local extrema at
f (a, b). For example, the above equations are true for the saddle
point indicated in the figure, which is not a local extremum.
Theorem 1 gives us necessary (but not sufficient) conditions for
f(a, b) to be a local extremum. Saddle point

Second Test for Local Extremum
The following theorem, using second derivatives, gives us
sufficient conditions for a critical point to produce a local
extremum or a saddle point.
Theorem: Assume that
1. z = f (x, y)
2. f x(a, b) = 0 and f y(a, b) = 0
3. All second-order partial derivatives of f exist in some
circular region containing (a, b) as a center
4. A = fxx(a, b), B = fxy(a, b), C = fyy(a, b)
(continued)

Second Test
(continued)
Then
Case 1. If AC - B2
> 0 and A < 0, then f (a, b) is a local
maximum.
Case 2. If AC - B2
> 0 and A > 0, then f (a, b) is a local
minimum.
Case 3. If AC - B2
< 0, then f (a, b) is a saddle point.
Case 4. If AC - B2
= 0, the test is inconclusive.

Procedure
There exists a three step procedure to assist you in finding
extrema.
Step 1. Find the critical point: Find (a, b) such that
f x(a, b) = 0 and f y(a, b) = 0 simultaneously.
Step 2. Compute A = fxx(a, b), B = fxy(a, b), C = fyy(a, b).
Step 3. Evaluate AC – B2
and classify the critical point using
theorem 2.

Example 1
Find local extrema for f (x, y) = 3 – x2
– y2
+ 6y
Step 1. Find the critical point:

Example 1
– y2
+ 6y
f x(x, y) = – 2x = 0 when x = 0
f y(x, y) = – 2y + 6 = 0 when y = 3.
There is a critical point at (0, 3).

Example 1
– y2
+ 6y
f x(x, y) = – 2x = 0 when x = 0
f y(x, y) = – 2y + 6 = 0 when y = 3.
A = – 2, B = 0, and C = – 2
and classify the critical point

Example 1
– y2
+ 6y
f x(x, y) = – 2x = 0 when x = 0
f y(x, y) = – 2y + 6 = 0 when y = 3.
A = – 2, B = 0, and C = – 2
and classify the critical point
AC – B2
= 4, A < 0. (0, 3) is a local maximum.

Example 2
The annual labor and automated equipment cost (in millions
of dollars) for producing TV sets is given by
C (x, y) = 2x2
+ 2xy + 3y2
– 16x – 18y + 54,
where x is the amount spent per year on labor, and y is the
amount spent per year on automated equipment (both in
millions of dollars).
Minimize the cost.

Example 2
(continued)
C (x, y) = 2x2
+ 2xy + 3y2
– 16x – 18y + 54
C x( x, y) = 4x + 2y – 16 = 0
C y( x, y) = 2x + 6y – 18 = 0
Solving these equations simultaneously gives the
critical point (3, 2).

Example 2
(continued)
Step 2. Compute A = Cxx(3, 2), B = Cxy(3, 2), C = Cyy(3, 2).
A = 4, B = 2, and C = 6
and classify the critical point.
AC – B2
= 20, A > 0. (3, 2) is a local minimum.
C x(x, y) = 4x + 2y – 16 = 0
C y(x, y) = 2x + 6y – 18 = 0

Example 2
(continued)
We found the critical value of the point (3, 2) yielded a
minimum cost so we need to calculate
C (3, 2) = 2 · 32
+ 2 · 3 · 2 + 3 · 22
- 16 · 3 - 18 · 2 + 54
= 12.
The minimum cost is 12 million dollars, when we spend
3 million dollars on labor and 2 million dollars on
automated equipment each year.

Summary
■ We learned how to use partial derivatives and the second-
derivative test to find local extrema.
■ We saw an application of this process in action.

bmb11e_ppt_15_3 lecture notes for students.ppt

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