*
Interpretation: When taking the first partial derivative of
f
(
x, y
) with respect to
x
, treat all instances of
the variable
y
(and any true constant) as a constant. Analogous statement with respect to
y
. See example
1 on p. 542.
*
Second order partials: see pp. 547548 for a discussion on second order partials.
*
Exercises: p.550 342
–
8.3 Maxima and Minima of Functions of Several Variables
*
Terms: relative maximum, relative maximum value, relative minimum, relative minimum value, absolute
maximum, absolute maximum value, absolute minimum, absolute minimum value, saddle point, critical
point, second derivative test.
*
Critical Point of
f
: A critical point of
f
is a point (
a, b
) in the domain of
f
such that both
∂f
∂x
(
a, b
) = 0 and
∂f
∂y
(
a, b
) = 0
or at least one of the partial derivative does not exist.
*
Determining Relative Extrema:
1. Find the critical points of
f
(
x, y
) by solving the system of simultaneous equations
f
x
= 0
f
y
= 0
2. The second derivative test for the critical point (
a, b
): Let
D
(
x, y
) =
f
xx
f
yy

f
2
xy
Then,
(a)
D
(
a, b
)
>
0 and
f
xx
(
a, b
)
<
0 implies that
f
(
x, y
) has a
relative maximum
at the point (
a, b
).
(b)
D
(
a, b
)
>
0 and
f
xx
(
a, b
)
>
0 implies that
f
(
x, y
) has a
relative minimum
at the point (
a, b
).
(c)
D
(
a, b
)
<
0 implies that
f
(
x, y
) has neither a rel. max. nor a rel. min. at the point (
a, b
). The
point (
a, b, f
(
a, b
)) is called a
saddle point
.
(d)
D
(
a, b
) = 0 implies that the test is inconclusive, so some other technique is needed to solve the
problem.
*
Exercises: pp.561563 130
–
8.5 Constrained Maxima and Minima and the Method of Lagrange Multipliers
*
Terms: unconstrained relative extremum, constrained relative extremum, method of Lagrange multipliers,
Lagrangian function
*
Substitution method: Want to find some extremum of a function
f
(
x, y
) subject to the constraint
g
(
x, y
) =
0. If
g
(
x, y
) = 0 is simple enough to solve for
x
or
y
, do so. For the sake of argument, say we could solve
for
y
and
y
=
r
(
x
) where
r
(
x
) is some expression in
x
. Then take the expression for
y
and plug it into
f
(
x, y
) giving
f
(
x, r
(
x
)).
So now you have a function of one variable
h
(
x
) =
f
(
x, r
(
x
)).
Now you can
use single variable methods to find the desired extremum. This is the same process if you solved for
x
instead. See example 1 on page 576 for a concrete example of this.
6