Microsoft Word - ECON2016 Unit 4 and 5 Practice Questions.docx
1
Units 4 and 5 Additional Practice Questions
Find additional practice questions for Units 4 and 5 in preparation for the graded quiz.
Solutions will not be posted but you can email me if you have any specific queries.
Unit 4 – Convexity and Concavity
Question 1
Classify the stationary point(s) of each of the following functions as convex, concave
(strictly or weakly) or neither.
(i)
(ii)
(iii)
(iv)
Question 2
Show that the Co
– Douglas production function , with
and is strictly concave for .
Unit 5 - Unconstrained and Constrained Optimization
Question 1
Find the global maximum and minimum values of each function and indicate where
they occur.
(i) subject to .
(ii) subject to .
(iii) subject to .
(iv) subject to .
Question 2
Find the stationary points and the values of the Lagrange multipliers for the
following problem.
(i) Min subject to
(ii) Max subject to
(iii) Max subject to and
Question 3
Determine whether the stationary point represents a maximum/minimum
subject to
Question 4
Dell, a manufacturer of computers and laptops, finds that it takes units of labour
and units of capital to produce units of the product. If a unit
of labour costs $100, unit of capital $200, and $200, 000 is budgeted for production,
2
2
2
12121 ),( xxxxxxf --=
21
2
22
2
XXXXXXXXXX),( xxxxxxxxf -+++=
21
2
2
2
121 32),( xxxxxxf ++=
12333),( 221
23
22
2
121 +--+= xxxxxxxf
( ) ba 2121, xAxxxf =
1,0
ba 1<+ ba 0,0 21
xx
2121 43),( xxxxf += 11,10 21 ££-££ xx
523),( 22
2
121 ++= xxxxf 102,100 21 ££££ xx
2
2
2
121 ),( xxxxf += 41,31 21 ££-££- xx
21
2
2
2
XXXXXXXXXX),( xxxxxxxxf +--+= 10,10 21 ££££ xx
f (x1, x2 ) = 4x1x2 + 7x1x3 + 9x2x3 x1x2x3 = 2016
f (x1, x2, x3) = 3x
2
1x2x3 x1 + x2 + x3 = 32
XXXXXXXXXX),,( xxxxxxf ++= 2
2
2
2
1 =+ xx 132 =+ xx
f (x1, x2 ) = 20x1 − 4x1
2 + 9x1x2 − 6x
2
2 +18x2 x1 +3x2 = 64
1x
2x f (x1, x2 ) =100x1
0.75x2
0.25
2
determine how many units should be expended on labour and how many units
should be expended on capital in order to maximize production. (You MUST justify
that the stationary point you found does indeed maximize the given function).
Question 5
Find the amount of capital (K) and labour (L) that should be employed to maximize
output (q) given the following Co
-Douglas production functions and constraints
set by PK, PL and budget B:
(You MUST justify that the stationary point you found does indeed maximize
the given function).
Question 6
Show that following problem is a convex programming problem and use the KKT
conditions to find the optimal solution.
Question 7
Consider the following nonlinear programming problem
Determine whether can be optimal by applying the KKT
conditions.
q(K,L) = K 0.4L0.6
PK = 8,PL = 6,B = 300
Maximize f (x1,x2 ) = 24x1 − x1
2 +10x2 − x2
2
subject to XXXXXXXXXXx1 ≤ 8
XXXXXXXXXXx2 ≤ 4
and XXXXXXXXXXx1 ≥ 0, x2 ≥ 0
0,0 XXXXXXXXXXand
3 XXXXXXXXXXsubject to
XXXXXXXXXXMaximize
21
21
3
22
3
1
2
11
³³
£+
-+-+=
xx
xx
xxxxxZ
( ) )2,1(, 21 =xx
Microsoft Word - Section_6_4_Lagrange.doc
Section 6.4 – Method of Lagrange Multipliers
237
Section 6.4
Method of Lagrange Multipliers
The Method of Lagrange Multipliers is a useful way to determine the minimum or maximum of a
surface subject to a constraint. It is an alternative to the method of substitution and works particularly
well for non-linear constraints. For the following examples, all surfaces will be denoted as ),( yxf
and all constraints as cyxg =),( .
The process follows an algorithm of steps which reveal another relationship between the input
variables that was neither given nor evident beforehand. We can use this relationship in conjunction
with the constraint to determine critical points of the surface on the constraint.
The Method of Lagrange Multipliers follows these steps:
1) Given a multivariable function ),( yxf and a constraint cyxg =),( , define the Lagrange
function to be )),((),(),( cyxgyxfyxL −−= λ , where λ (lambda) is multiplied
(distributed) through the constraint portion.
2) Determine the partial derivatives xL and yL .
3) Set the partial derivatives xL and yL equal to zero.
4) Make note of any immediate solutions for x and y that satisfy 0=xL and 0=yL . Often
there are none, so proceed to step 5.
5) Isolate the λ in each equation.
6) Equate the two λ equations, dropping out the λ altogether.
7) Reduce the equation from step 6 as far as possible. This is the relationship between x and
y that shall be substituted into the constraint.
8) Substitute this equation into the constraint and alge
aically solve for the variable that
emains. This is where the critical points start to be determined.
9) Solve for the other variable by re-evaluating your results in step 8 back into the constraint.
If just one critical point results, you cannot necessarily determine if it is a minimum or maximum just
y itself, so you must study the shape of the surface and make inferences based on its shape and the
elative location of the constraint (this is where a contour map helps). If two or more points result,
then minima and maxima are easily determined by comparing their z-values. Some points may be
neither minima nor maxima and are then ignored.
In step 7 above, it is possible to come to an impossible (unsolvable) alge
aic equation, in which case
we must then refer to the possible solutions for x or y that were noted in step 4, and try them directly
into the constraint to determine the critical points. This is shown in Example 5 below. Often in a case
like this, it is just easier to resort to direct substitution and deal with the alge
a from the start.
Section 6.4 – Method of Lagrange Multipliers
238
The examples
eak down into the following scenarios:
• Example 1 features a linear constraint, and illustrates both methods—Lagrange and
substitution—for locating its critical point for comparison’s sake.
• Example 2 features a non-linear (circular, in this case) constraint and is more ‘typical’ of a
scenario in which Lagrange is the prefe
ed method over substitution.
• Example 3 shows a situation where more than one minimum and/or maximum point can be
found. This happens usually on highly symmetrical surfaces.
• Example 4 is a case where the method works just as it should, yet comes to a difficult
higher-degree (4th degree in this case) polynomial when the λ are equated.
• Example 5 shows how to handle the case when equating the λ expressions leads to an
unsolvable equation.
Example 1: Determine the minimum and maximum values on yxyxyxf 22),( 22 −−+= subject to
the constraint 42 =+ yx .
Discussion: It helps to have an idea of the shape of the surface and how the constraint will appear
when viewed as part of the surface. The surface ),( yxf is a paraboloid with its vertex at )2,1,1( − .
The paraboloid opens upward making the vertex the absolute minimum point on the surface. The
constraint is a straight line when viewed on the xy-plane contour plot. When conformed onto the
surface itself, it is a parabolic cross-section of the surface. Since the surface opens up, any linear
cross-section will also be a parabola opening up, so it stands to reason that the surface will have a
minimum point but no maximum point subject to the constraint.
Figure. Contour map with constraint path for Example 1.
Section 6.4 – Method of Lagrange Multipliers
239
Form the Lagrange Function first, following the form )),((),(),( cyxgyxfyxL −−= λ :
λλλ
λ
4222),(
)42(22),(
22
22
+−−−−+=⇒
−+−−−+=
yxyxyxyxL
yxyxyxyxL
Next, partially differentiate ),( yxL with respect to x and y, and set each equation equal to zero:
0222
022
222
22
=−−
=−−
⇒
−−=
−−=
λ
λ
λ
λ
y
x
yL
xL
y
x
There are no immediate solutions to x or y that satisfy the above two equations, so we proceed and set
the two partial derivatives to zero and isolate the λ in each:
1
22
0222
022
2
22 −==
−=
⇒
=−−
=−−
− y
x
y
x
yλ
λ
λ
λ
By transitivity, the two expressions for λ can be equated to one another and reduced:
12122 −=⇒−=− xyyx
The result is a relationship between x and y that we can now insert into the constraint by substitution:
5
6
65424
XXXXXXXXXX
=⇒
=⇒=−+⇒
=−+⇒=+
x
xxx
xxyx
We now have the x-coordinate of the minimum point along the constraint. To find y, back-substitute
this value into the constraint:
5
7
10
14
5
14
5
XXXXXXXXXX ==⇒=⇒=+⇒=+ yyyyx .
Thus, the surface has a minimum at ),,( 5
9
5
7
5
6 − , where 595756 ),( −== fz .
Since the constraint above was linear, this same problem could have been solved by direct
substitution: in the constraint 42 =+ yx , solve for x to get yx 24 −= and substitute into the function
),( yxf :
yyyyyyf XXXXXXXXXX(),24( 22 −−−+−=− .
Section 6.4 – Method of Lagrange Multipliers
240
Now just work out the alge
a by expanding then collecting terms:
8145
XXXXXXXXXX
XXXXXXXXXX(
XXXXXXXXXX(),24(
2
22
2
22
+−=
−+−++−=
−−−+−−=
−−−+−=−
yy
yyyyy
yyyyy
yyyyyyf
Thus, the cross-section formed by the linear constraint “cutting through” the paraboloid is a parabola
with an equation 8145),24( 2 +−=− yyyyf . To find its critical point, differentiate and set equal to
zero, as usual:
5
7
10
XXXXXXXXXX ==⇒=−⇒−=′ yyyf .
Obtain x by substituting 5
7=y into the constraint: XXXXXXXXXX =−=x . Finally, obtain z by evaluating
oth x and y into the original function. ��
Figure. Example 1, with minimum point on the constraint identified.
Comment on Example 1: