Texas Instruments TI-84 Plus Manual page 111

TI-83, TI-83 Plus, TI-84 Plus Guide
strained optimization problem with the functions given in Example 1 of Section 10.3 – the
Cobb-Douglas production function f ( L, K ) = 48.1 L
8 L + K = 98 where L worker hours (in thousands) and $ K thousand capital investment are for a
mattress manufacturing process.
We first find the critical point(s). Because this function does not yield a linear system of
partial derivative equations, we use the algebraic method. We employ a slightly different order
of solution than that shown in the text. The system of partial derivative equations is
−
0.4
28.86 L
0.6
19.24 L
K
8 L + K = 98
Equation 4 was derived by substituting λ from the second equation on the left into the first,
and equation 5 was derived by solving the third equation on the left for K . We now solve this
system of 2 equations (equations 4 and 5) in 2 unknowns ( L and K ) using the methods shown
on page 106.
Clear the list. Enter the function f
Y=
in and the expression for K in
Y1
Rewrite the other equation (equation 4)
so that it equals 0, and enter the non-
zero side in
Y3
(See the note below.)
NOTE: Remember that
Do the same for the other
equation that equals 0 and contains only one variable, namely
many answers there are to this equation.)
Use the
SOLVER
= 0. Try several different guesses
Y3
and see that they all result in the same
solution.
The calculator stores the value of
Now return to the home screen and
evaluate Y2. Store this value in K.
to display the value of f at this
Y1
point.
Classifying critical points when a constraint is involved is done by graphing the constraint
on a contour graph or by examining close points. Your calculator cannot help with the contour
graph classification – it must be done by hand. We illustrate the procedure used to examine
close points for this Cobb-Douglas production function .
We now test close points to see if this output value of f is maximum or minimum. Remem-
ber that whatever close points you choose, they must be near the critical point and they must be
on the constraint g .
WARNING: Do not round during this procedure. Rounding of intermediate calculations
and/or inputs can give a false result when the close point is very near to the optimal point.
Copyright © Houghton Mifflin Company. All rights reserved.
0.4
= 8 λ
K
⇒
28.86 L
−
0.6
= λ
⇒
K = 98 – 8 L
.
Y2
. Substitute into .
Y2 Y3
= . Put the cursor on the first
K Y2
in . The expression now in
K Y3
to solve the equation
L.
Enter
0.6 0.4
subject to the constraint g ( L, K ) =
K
−
0.4 0.4 0.6
= 8(19.24 L
K K
in
K
is the left-hand side of an
Y3
. (We are not sure how
L
−
0.6
)
and replace by
Y3 K Y2
[4]
[5]
.
111