Finding Critical Points Using Matrices - Texas Instruments TI-84 Plus Manual

TI-83, TI-83 Plus, TI-84 Plus Guide
Take the derivative of
, the partial derivative of V with respect to h and then h
V
hh
again. Find V
Take the derivative of
, the partial derivative of V with respect to w and then h.
V
wh
−
Find V = 36 at the critical point.
wh
Take the derivative of
, the partial derivative of V with respect to h and then w.
V
hw
−
Find V = 36 at the critical point. (Note: V
hw
Take the derivative of
, the partial derivative of V with respect to w and then w
V
ww
again. Find V
The second partials matrix is
− −
D = ( 72)( 72) – (
the critical point, the Determinant Test tells us that (h, w, V) =
(18, 18, 11664) is a relative maximum point.
NOTE: The values of the second partial derivatives were not very difficult to determine with-
out the calculator in this example. However, with a more complicated function, we strongly
suggest using the above methods to provide a check on your analytic work to avoid making
simple mistakes. You can also use the function in
We use some randomly chosen values of
the values at the critical point, because the critical point values give 0 for
Store different values in
evaluate at these values. Then
Y2
evaluate the calculator's derivative of
with respect to
Y1
values.
Evaluate at these stored values of
Y3
and . Then evaluate the
H W
calculator's derivative of
respect to at these same values.
W
10.2.2 FINDING CRITICAL POINTS USING MATRICES To find the critical point(s) of a
smooth, continuous multivariable function, we first find the partial derivatives with respect
to each of the input variables and then set the partial derivatives equal to zero. This gives a
system of equations that needs to be solved. Here we illustrate solving this system using an
orderly array of numbers that is called a matrix.
WARNING: The matrix solution method applies only to linear systems of equations. That
is, the system of equations should not have any variable appearing to a power higher than 1
and should not contain a product of any variables. If the system is not linear, you must use
algebraic solution methods to solve the system. Using your calculator with the algebraic
solution method is illustrated on page 106 of this Guide.
Copyright © Houghton Mifflin Company. All rights reserved.
with respect to h and we have
Y2 =
V
h
−
= 72 at the critical point.
hh
with respect to h and we have
Y3 =
V
w
with respect to w and we have
Y2 =
V
h
with respect to w and we have
Y3 =
V
w
−
= 72 at the critical point.
ww
L
N M V V
hh
V V
wh
−
36) 2 = 3888. Because D > 0 and V
and and
H W
at these same
H
with
Y1
must equal V
wh hw
O
Q P
hw
. Find the value of
ww
< 0 at
hh
to check the derivatives in
Y1
and to illustrate this quick check rather than use
H W
If you prefer, enter the value
of H in the 3rd position.
.)
If you prefer, enter the value
of W in the 3rd position.
and
Y2 Y3
and .
Y2 Y3
.
107